The answer is (a) The time taken to collide is 6.52 s (b) The distance covered by the first player before the collision is 11.36 m.
Given that Two soccer players start from rest, 40 m apart.
They run directly toward each other, both players accelerating.
The first player's acceleration has a magnitude of 0.47 m/s2.
The second player's acceleration has a magnitude of 0.47 m/s2.
(a) To find time of collision
The equation of motion for the two players are:
First player's distance x1= 1/2 a1t^2
Second player's distance x2= 40m - 1/2 a2t^2 where x1 = x2
When the players collide Time taken to collide is the same for both players 0.5 a1t^2 = 40m - 0.5 a2t^2.5 t^2(a1+a2) = 40m.t^2 = 40m/0.94 = 42.55 m
Seconds passed for the collision to take place = √t^2 = 6.52s
(b) How far has the first player run?
First player's distance x1= 1/2 a1t^2= 1/2 x 0.47m/s^2 x (6.52s)^2= 11.36m
Therefore, the first player ran 11.36m before the collision.
Hence the required answer is: (a) The time taken to collide is 6.52 s (b) The distance covered by the first player before the collision is 11.36 m.
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A 20 gram hollow sphere rolls down a 25 cm high ramp from rest. The sphere has a radius of 1.5 cm. You can ignore air resistance. What is the sphere's linear speed at the bottom of the ramp? 3.46 m/s 0.87 m/s 1.73 m/s 4.65 m/s 2.05 m/s 1.34 m/s
The linear speed of a hollow sphere that rolls down a 25 cm high ramp from rest can be determined as follows:
Given data: mass of the sphere (m) = 20 g = 0.02 kg
The radius of the sphere (r) = 1.5 cm = 0.015 m
height of the ramp (h) = 25 cm = 0.25 m
Acceleration due to gravity (g) = 9.81 m/s².
Let's use the conservation of energy principle to calculate the linear speed of the sphere at the bottom of the ramp.
The initial potential energy (U₁) is given by: U₁ = mgh where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height of the ramp.
U₁ = 0.02 kg × 9.81 m/s² × 0.25 m = 0.049 J.
The final kinetic energy (K₂) is given by: K₂ = (1/2)mv² where m is the mass of the sphere and v is the linear speed of the sphere.
K₂ = (1/2) × 0.02 kg × v².
Let's equate the initial potential energy to the final kinetic energy, that is:
U₁ = K₂0.049 = (1/2) × 0.02 kg × v²0.049
= 0.01v²v² = 4.9v = √(4.9) = 2.21 m/s (rounded to two decimal places).
Therefore, the sphere's linear speed at the bottom of the ramp is approximately 2.21 m/s.
Hence, the closest option (d) to this answer is 2.05 m/s.
The sphere's linear speed is 2.05 m/s.
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Show that whenever white light is passed through a diffraction grating of any spacing size, the violet end of the spectrum in the third order on a screen always overlaps the red end of the spectrum in the second order.
When the white light passes through the diffraction grating, the violet light will be deviated at a larger angle than the red light. This causes the violet light to overlap with the red light on the screen, as the violet light has a wider spread due to its larger angle of diffraction.
When white light passes through a diffraction grating, it undergoes diffraction, which causes the different colors of light to spread out. This creates a pattern of colored bands known as a spectrum. The spacing of the grating determines the angles at which different orders of the spectrum are observed on a screen.
To understand why the violet end of the spectrum in the third order overlaps with the red end of the spectrum in the second order, we need to consider the relationship between the angles of diffraction for different colors.
The angle at which a specific color is diffracted depends on its wavelength. The violet end of the spectrum has a shorter wavelength than the red end. Since the third order is associated with a higher angle of diffraction than the second order, we can deduce that the violet light will be diffracted at a larger angle than the red light.
As a result, when the white light passes through the diffraction grating, the violet light will be deviated at a larger angle than the red light. This causes the violet light to overlap with the red light on the screen, as the violet light has a wider spread due to its larger angle of diffraction.
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: c. List three materials that was used during effect of concentration experiment. (1.5 marks - 0.5 mark each) Question 2:(5.0 marks) a. List three unknown metals that was used during the flame test. (1.5 mark - 0.5 mark each) b. What base was used doing titration experiment shown to you. (1.0 mark) c. What acid was used doing titration experiment shown to you. (1.0 nark)
c. For the effect of concentration experiment, three materials commonly used are:
1. Beakers or test tubes: These containers are used to hold the solutions of varying concentrations.
2. Measuring cylinders or pipettes: These tools are used to accurately measure the volumes of solutions needed for the experiment.
3. Stirring rods or magnetic stirrers: These are used to mix the solutions thoroughly and ensure homogeneity.
a. In the flame test, three unknown metals were used to observe their characteristic flame colors:
1. Sodium: Sodium typically produces a yellow-orange flame color.
2. Copper: Copper usually produces a blue-green flame color.
3. Potassium: Potassium often produces a lilac or lavender flame color.
b. The base used in the titration experiment depends on the specific experiment being conducted. Without further information, it is not possible to determine the specific base used.
c. Similarly, the acid used in the titration experiment would depend on the nature of the experiment. Without additional information, it is not possible to determine the specific acid used.
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4. Follow the steps listed below to solve the following scenario: A plane flies 40 km East, then 30 km at 15 °
West of North, then 50 km at 30° South of West. What is its displacement (resultant) vector?
a. Assign a letter ("A", "B", "C", etc.) to each vector. Record the magnitudes and the angles of each
vector into your lab book.
b. Write an addition equation for your vectors. For example: A + B + C = R
c. Find the resultant vector by adding the vectors graphically:
i. Draw a Cartesian coordinate system.
ii. Determine the scale you want to use and record it (example: 1 cm=10 km).
iii. Add the vectors by drawing them tip-to-tail. Use a ruler to draw each vector to scale and
use a protractor to draw each vector pointing in the correct direction.
iv. Label each vector with the appropriate letter, magnitude, and angle. Make sure that the
arrows are clearly shown.
v. Draw the resultant vector.
vi. Use the ruler to determine the magnitude of the resultant vector. Show your calculation,
record the result, and draw a box around it. Label the resultant vector on your diagram.
vii. Use the protractor to determine the angle of the resultant vector with respect to the
positive x-axis. Record the value and draw a box around it. Label this angle on your diagram.
d. Find the resultant vector by adding the vectors using the analytical method:
i. Calculate the x and y-components of each vector.
ii. Find the x-component and the y-component of the resultant vector.
iii. Find the magnitude of the resultant vector. Draw a box around your answer.
iv. Find the angle that the resultant makes with the positive x-axis. Draw a box around your
answer.
e. Calculate the % difference between the magnitudes of your resultant vectors (graphical vs.
analytical).
f. Compare your two angles (measured vs. calculated).
a. Magnitudes and angles of each vector:
A: 40 km (East), B: 30 km (15° West of North), C: 50 km (30° South of West).
b. Addition equation: A + B + C = R.
c. Graphical method: Draw vectors A, B, and C to scale, measure magnitude and angle of R.
d. Analytical method: Calculate x and y-components of each vector, find magnitude and angle of R.
e. % difference between graphical and analytical magnitudes of R.
f. Comparison of measured and calculated angles of R.
To solve the scenario, follow these steps:
a. Assign letters and record magnitudes and angles:
Let A be the vector representing the plane flying 40 km East, B be the vector for 30 km at 15° West of North, and C represent 50 km at 30° South of West.
A: Magnitude = 40 km, Angle = 0° (East)
B: Magnitude = 30 km, Angle = 75° (15° West of North)
C: Magnitude = 50 km, Angle = 240° (30° South of West)
b. Write the addition equation: A + B + C = R
c. Find the resultant vector graphically:
- Draw a Cartesian coordinate system.
- Determine the scale (e.g., 1 cm = 10 km).
- Draw vectors A, B, and C to scale, tip-to-tail.
- Label each vector with letter, magnitude, and angle.
- Draw the resultant vector R.
- Measure the magnitude of R using a ruler and record it.
- Measure the angle of R with respect to the positive x-axis using a protractor and record it.
d. Find the resultant vector analytically:
- Calculate x and y-components of each vector.
- Find the x and y-components of R.
- Calculate the magnitude of R and record it.
- Determine the angle of R with the positive x-axis and record it.
e. Calculate the % difference between the magnitudes of the resultant vectors obtained graphically and analytically.
f. Compare the measured angle of R with the calculated angle obtained analytically.
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The propagation of uncertainty formula for the equation y=mx+b is (∂m∂yδm)2+(∂x∂yδx)2+(∂b∂yδb)2 where for example δm is the uncertainty on m and ∂m∂y is the partial derivative of y with respect to m. If m=0.4+1−0.9⋅x=−0.7+/−0.1 and b=−3.9+/−0.6 then what is the uncertainty on y QUESTION 6 Find the uncertainty in kinetic energy. Kinetic energy depends on mass and velocity according to this function E(m,v)=1/2mv2. Your measured mass and velocity have the following uncertainties δm=0.47 kg and δV=1.05 m/s. What is is the uncertainty in energy, δE, if the measured mass, m=4.55 kg and the measured velocity, v= −0.32 m/s ? Units are not needed in your answer.
The uncertainty on y is 0.392.The formula for kinetic energy is E(m,v)=1/2mv^2. The propagation of uncertainty formula for the equation y=mx+b is given by:
(∂m/∂y * δm)^2 + (∂x/∂y * δx)^2 + (∂b/∂y * δb)^2
where δm is the uncertainty on m and ∂m/∂y is the partial derivative of y with respect to m, δx is the uncertainty on x and ∂x/∂y is the partial derivative of y with respect to x, and δb is the uncertainty on b and ∂b/∂y is the partial derivative of y with respect to b.
Given that m=0.4+1−0.9⋅x=−0.7+/−0.1 and b=−3.9+/−0.6, the uncertainty on y can be found by substituting the values in the above formula.
(∂m/∂y * δm)^2 + (∂x/∂y * δx)^2 + (∂b/∂y * δb)^2
= (∂(0.4+1−0.9⋅x−3.9)/∂y * δm)^2 + (∂(0.4+1−0.9⋅x−3.9)/∂y * δx)^2 + (∂(0.4+1−0.9⋅x−3.9)/∂y * δb)^2
= (-0.9 * δm)^2 + (-0.9 * δx)^2 + δb^2
= (0.81 * 0.1^2) + (0.81 * 0.1^2) + 0.6^2
= 0.0162 + 0.0162 + 0.36
= 0.392
The uncertainty in energy δE can be found by using the formula:
(∂E/∂m * δm)^2 + (∂E/∂v * δv)^2
= (1/2 * v^2 * δm)^2 + (mv * δv)^2
= (1/2 * (-0.32)^2 * 0.47)^2 + (4.55 * (-0.32) * 1.05)^2
= 0.0192 + 2.1864
= 2.2056
Thus, the uncertainty in energy δE is 2.2056.
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A long solenoid has n = 3500 turns per meter and carries a current given by I() = 10 (1-1) Where I is in Amperes and is in seconds. Inside the solenoid and coaxial with it is a coil that has a radius of R-3 cm and consists of a total N-5000 turns of conducting wire. #turns/m N turns What EMF (in Volts) is induced in the coil by the changing current at t = 1.1 s?
The induced EMF in the coil at t = 1.1 s is 1.1 V. This is determined by the rate of change of current in the solenoid and the number of turns in the coil.
The EMF induced in a coil is given by the equation EMF = -N * dΦ/dt, where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux through the coil.
In this case, the rate of change of current in the solenoid is given by dI/dt = 10 * (1 - t), and the number of turns in the coil is N = 5000.
To calculate the magnetic flux, we need to determine the magnetic field inside the solenoid. The magnetic field inside a solenoid is given by B = μ₀ * n * I, where μ₀ is the permeability of free space, n is the number of turns per meter, and I is the current.
Substituting the values into the equation, we get B = (4π * 10^(-7) * 3500 * 10 * (1 - t)) T.
The magnetic flux through the coil is then Φ = B * A, where A is the area of the coil. Since the coil is coaxial with the solenoid, the area is given by A = π * R².
Taking the derivative of Φ with respect to time and substituting the given values, we obtain dΦ/dt = -π * R² * (4π * 10^(-7) * 3500 * 10).
Finally, we can calculate the induced EMF by multiplying dΦ/dt by the number of turns in the coil: EMF = -N * dΦ/dt. Plugging in the values, we find that the induced EMF at t = 1.1 s is 1.1 V.
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Question 6 A horizontal 16-N force is needed to slide a 50-kg box across a flat surface at a constant velocity of 3.5 ms. What the coefficent of anec friction between the band the O 0.10 O 011 0 0.13
The coefficient of static friction between the box and surface, given that a 16 N force is needed is 0.03
How do i determine the coefficient of static friction?First, we shall obtain the normal reaction. Details below:
Mass of object (m) = 50 KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = ?N = mg
= 50 × 9.8
= 490 N
Finally, we shall obtain the coefficient of static friction. Details below:
Force needed = 16 NNormal reaction (N) = 490 NCoefficient of friction (μ) =?μ = F / N
= 16 / 490
= 0.03
Thus, we can conclude that the coefficient of friction is 0.03. None of the options are correct
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QUESTION 2-ANSWER ALL PARTS (a) A pump is used to abstract water from a river to a water treatment works 20 m above the river. The pipeline used is 300 m long, 0.3 m in diameter with a friction factor A of 0.04. The local headloss coefficient in the pipeline is 10. If the pump provides 30 m of head Determine the (i) pipeline flow rate. (ii) local headloss coefficient of the pipeline, if the friction factor is reduced to A=0.01. Assume that the flow rate remains the same as in part i) and that the other pipe properties did not change. [10 marks]
Pump is used to abstract water from a river to a water treatment plant 20 m above the river. The pipeline used is 300 m long, 0.3 m in diameter with a friction factor A of 0.04. K = 19.6, K' = 10408.5
The pipeline flow rate and local headloss coefficient can be calculated as follows;
i) Pipeline Flow rate:
Head at inlet = 0
Head at outlet = 20 + 30 = 50m
Frictional loss = f x (l/d) x (v^2/2g)
= 0.04 x (300/0.3) x (v^2/2 x 9.81)
= 39.2 x v^2x v
= (Head at inlet - Head at outlet - Frictional Loss)^0.5
= (0 - 50 - 39.2v^2)^0.5Q
= A x v
= πd^2/4 x v
= π(0.3)^2/4 x (0.27)^0.5
= 0.0321 m3/s
= 32.1 L/s
ii) Local Headloss Coefficient:
Frictional Loss = f x (l/d) x (v^2/2g)
= 0.01 x (300/0.3) x (v^2/2 x 9.81)
= 9.8 x v^2Head at inlet
= 0Head at outlet
= 50 + 30 = 80m
Total Headloss = Head at inlet - Head at outlet
= 0 - 80
= -80 m
Since the flow rate remains the same, Q = 0.0321 m3/s
Frictional Loss = f x (l/d) x (v^2/2g)
= K x (v^2/2g)
= K' x Q^2 (K' = K x d^5 / l g)^0.5
= 9.8 x v^2
= K x (v^2/2g)
= K' x Q^2
Hence, K = 19.6, K' = 10408.5
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A large punch bowl holds 3.50 kg of lemonade (which is essentially water) at 22.0 °C. A 5.90x10-2-kg ice cube at -15.0 °C is placed in the lemonade. You may want to review (Pages 607 - 608). Part A What is the final temperature of the system?
the final temperature of the system is approximately 11.29 °C.
calculate the heat gained by the ice cube using the equation:
Q = m * c * ΔT
where,
Q = is the heat gained by ice
m = is the mass of the ice cube
c = is the specific heat capacity of ice,
ΔT = is the change in temperature of the ice.
Given:
m = 5.90x[tex]10^-2[/tex] kg
c = 2100 J/kg°C (specific heat capacity of ice)
ΔT = t- (-15.0 °C) = t + 15.0 °C (final temperature of the ice cube is t)
Now, calculate the heat lost by the lemonade using the equation:
Q = m * c * ΔT
where,
Q = is the heat lost of lemonade
m= is the mass of the lemonade
c = is the specific heat capacity of water,
ΔT= is the change in temperature of the lemonade.
Given:
m = 3.50 kg
c = 4186 J/kg°C (specific heat capacity of water)
ΔT= t - 22.0 °C (final temperature of the lemonade is t)
Since there is no heat exchange with the surroundings, the heat gained by the ice cube is equal to the heat lost by the lemonade:
Q of ice = Q of lemonade
m * c * ΔT= m * c * ΔT
Substituting the given values, we can solve for t:
(5.90x[tex]10^-2[/tex]kg) * (2100 J/kg°C) * (t + 15.0 °C) = (3.50 kg) * (4186 J/kg°C) * (t - 22.0 °C)
simplifying equation:
0.1239 kg J/°C * t + 0.1239 kg J = 14.651 kg J/°C * t - 162.872 kg J
-14.5271 kg J/°C * t = -163.9959 kg J
divide both sides by -14.5271 kg J/°C to solve for t:
t = (-163.9959 kg J) / (-14.5271 kg J/°C)
t ≈ 11.29 °C
Therefore, the final temperature of the system is approximately 11.29 °C.
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Physical Science
Given the Lewis symbols for carbon and oxygen below, draw the Lewis structure of CO2 (carbon dioxide). Remember to indicate single, double or triple bonds where appropriate.
Carbon:
Oxygen:
The Lewis structure of CO2 (carbon dioxide) is as follows:
O=C=O
To draw the Lewis structure of CO2, we follow these steps:
1. Determine the total number of valence electrons: Carbon (C) has 4 valence electrons, and each oxygen (O) atom has 6 valence electrons. Since we have two oxygen atoms, the total number of valence electrons is 4 + 2(6) = 16.
2. Write the skeletal structure: Carbon is the central atom in CO2. Place the carbon atom in the center and arrange the oxygen atoms on either side.
O=C=O
3. Distribute the remaining electrons: Distribute the remaining 16 valence electrons around the atoms to fulfill the octet rule. Start by placing two electrons between each atom as a bonding pair.
O=C=O
4. Complete the octets: Add lone pairs of electrons to each oxygen atom to complete their octets.
O=C=O
5. Check for octet rule and adjust: Check if all atoms have fulfilled the octet rule. In this case, each atom has a complete octet, and the structure is correct.
The final Lewis structure for carbon dioxide (CO2) is shown above, where the lines represent the bonding pairs of electrons.
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Explain how stellar evolution, and the universe would be
different if carbon was the most bound element instead of Iron.
If carbon were the most bound element instead of iron, stellar evolution and the universe would be significantly different. Carbon-based life forms would be more common, and the formation of heavy elements through stellar nucleosynthesis would be altered.
If carbon were the most bound element instead of iron, several implications would arise:
Stellar Evolution: Carbon fusion would become the primary process in stellar nucleosynthesis, leading to a different sequence of stellar evolution. Stars would undergo carbon burning, producing heavier elements and releasing energy.
The life cycle of stars, their sizes, lifetimes, and eventual fates would be modified.
Abundance of Carbon:
Carbon-based molecules, essential for life as we know it, would be more prevalent throughout the universe.
Carbon-rich environments would be more common, potentially supporting a wider range of organic chemistry and the development of carbon-based life forms.
Element Formation: The synthesis of heavier elements through stellar nucleosynthesis would be affected.
Iron is a crucial element for the formation of heavy elements through processes like supernova explosions. If carbon were the most bound element, alternative mechanisms for heavy element formation would emerge, potentially leading to a different abundance and distribution of elements in the universe.
Overall, the universe's composition, the prevalence of carbon-based life, and the processes involved in stellar evolution and element formation would be significantly different if carbon were the most bound element instead of iron.
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billy, a student, sounds two tuning forks that are supposed to be tuned to A 440.0hz. in which one is correct. When sounded with the other tuning ford, he hears a periodic volume change at a rate of 24 times in 6.0s
a) In physics, what is this called?
b) What would be the possible frequencies for the tuning fork that happens to be out of tune?
In physics, the periodic volume change heard when two sound waves with nearly similar frequencies interfere with each other is called beats. The frequency of the out-of-tune tuning fork is 222 Hz.
When two sound waves interfere with each other, the periodic volume change heard when two sound waves with nearly similar frequencies interfere with each other is called beats.
The frequency of the out-of-tune tuning fork can be calculated from the number of beats heard in a given time. Billy hears 24 beats in 6.0 seconds. Therefore, the frequency of the out of tune tuning fork is 24 cycles / 6.0 seconds = 4 cycles per second.
In one cycle, there are two sounds: one of the tuning fork, which is at a frequency of 440.0 Hz, and the other is at the frequency of the out-of-tune tuning fork (f). The frequency of the out-of-tune tuning fork can be calculated by the formula; frequency of the out-of-tune tuning fork (f) = (Beats per second + 440 Hz) / 2.
Substituting the values, we get;
frequency of the out-of-tune tuning fork (f) = (4 Hz + 440 Hz) / 2 = 222 Hz.
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(d) If a battery contains 2500 milliAmp-hours (mAh) of charge, how much total energy can it deliver while operating an electrical device at Z volts?
The total energy a battery can deliver at voltage Z is 2.5 Ah multiplied by Z.
To calculate the total energy that a battery can deliver while operating an electrical device at a specific voltage (Z), we need to convert the charge capacity of the battery from milliamp-hours (mAh) to amp-hours (Ah) and then multiply it by the voltage.
1. Convert the charge capacity from milliamp-hours (mAh) to amp-hours (Ah):
Divide the given charge capacity (2500 mAh) by 1000 to convert it to amp-hours:
2500 mAh / 1000 = 2.5 Ah
2. Calculate the total energy:
Multiply the charge capacity in amp-hours (2.5 Ah) by the voltage (Z):
Total Energy = Charge Capacity (Ah) × Voltage (Z)
Total Energy = 2.5 Ah × Z
Therefore, the total energy the battery can deliver while operating an electrical device at Z volts is 2.5 Ah × Z.
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what must be the radius (in cm) of a disk of mass 9kg, so that it
has the same rotational inertia as a solid sphere of mass 5g and
radius 7m?
Give your answer to two decimal places
The radius (in cm) of a disk of mass 9kg, so that it has the same rotational inertia as a solid sphere of mass 5g and radius 7m should be 6.13 cm (approximately).
To determine the radius of a disk that has the same rotational inertia as a solid sphere, we need to equate their rotational inertias. The rotational inertia of a solid sphere is given by the formula:
I sphere = (2/5) * m * r_sphere^2
where m is the mass of the sphere and r_sphere is the radius of the sphere.
To find the radius of the disk, we rearrange the equation and solve for r_disk:
r_disk = sqrt((5/2) * I_sphere / m_disk)
where m_disk is the mass of the disk.
Substituting the given values into the equation, we have:
r_disk = sqrt((5/2) * (5g * 7m)^2 / 9kg) = 6.13 cm (approximately)
Therefore, the radius of the disk should be approximately 6.13 cm to have the same rotational inertia as the given solid sphere.
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The radius (in cm) of a disk of mass 9kg, so that it has the same rotational inertia as a solid sphere of mass 5g and radius 7m should be 6.13 cm (approximately).
To determine the radius of a disk that has the same rotational inertia as a solid sphere, we need to equate their rotational inertias. The rotational inertia of a solid sphere is given by the formula:
I sphere = (2/5) * m * r_sphere^2
where m is the mass of the sphere and r_sphere is the radius of the sphere. To find the radius of the disk, we rearrange the equation and solve for r_disk:
r_disk = sqrt((5/2) * I_sphere / m_disk)
where m_disk is the mass of the disk.
Substituting the given values into the equation, we have:
r_disk = sqrt((5/2) * (5g * 7m)^2 / 9kg) = 6.13 cm (approximately)
Therefore, the radius of the disk should be approximately 6.13 cm to have the same rotational inertia as the given solid sphere.
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ADVD disc has a radius 6.0 cm and mass 28 gram. The moment of inertia of the disc is % MR2 where M is the mass, R is the radius. While playing music, the angular velocity of the DVD is 160.0 rad/s. Calculate [a] the angular momentum of the disc [b] While stops playing, it takes 2.5 minutes to stop rotating. Calculate the angular deceleration. [C] Also calculate the torque that stops the disc.
Given that,Radius of the ADVDisc, r = 6.0 cm = 0.06 m
Mass of the disc, M = 28 g = 0.028 kg
Moment of Inertia of the disc,
I = MR² = 0.028 × 0.06² = 0.00010 kg m²
Angular Velocity, ω = 160.0 rad/s[a]
Angular Momentum, L = Iω= 0.00010 × 160.0 = 0.016 Nm s[b]
Angular deceleration, α = -ω/t, where t = 2.5 min = 150 sα = -160/150 = -1.07 rad/s²
[Negative sign indicates deceleration][c] Torque that stops the disc is given by,Torque = I αTorque = 0.00010 × (-1.07) = -1.07 × 10⁻⁵ NmAns:
Angular momentum of the disc, L = 0.016 Nm s;Angular deceleration, α = -1.07 rad/s²;Torque that stops the disc = -1.07 × 10⁻⁵ Nm.
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A block is held stationary on a ramp by the frictional force on it from the ramp. A force F with arrow, directed down the ramp, is then applied to the block and gradually increased in magnitude from zero. As the magnitude of F with arrow is increased from zero, what happens to the direction and magnitude of the frictional force on the block?
The direction and initial magnitude of the frictional force on the block will not change as the force F applied on the block progressively increases from zero.
When the block is at rest, the force of friction opposes the force that tends to slide the block down the ramp because it acts in the direction opposite to the motion or tendency of motion. However, as soon as the applied force F exceeds the maximum static frictional force, the block will start to move. At this point, kinetic friction replaces static friction as the dominant type of friction. The kinetic friction force usually has a smaller magnitude than the maximum static friction force.
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As the magnitude of the force F directed down the ramp is increased from zero, the direction of the frictional force on the block stays the same.
However, the magnitude of the frictional force decreases to match the magnitude of the applied force until the block begins to slide. Once the block begins to slide, the magnitude of the frictional force remains constant at the sliding friction force magnitude. Additionally, the direction of the sliding frictional force is opposite to the direction of the block's motion. This is consistent with Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. Therefore, the force of the block on the ramp is met with a force of the ramp on the block that is opposite in direction and equal in magnitude, up until the point where the block begins to slide down the ramp. After this point, the magnitude of the frictional force will remain constant, as the block slides down the ramp.
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A 8.0μF capacitor, a 11μF capacitor, and a 14 Part A uF capacitor are connected in parallel. What is their equivalent capacitance? Express your answer in microfarads.
When capacitors are connected in parallel, their equivalent capacitance can be obtained using the formula below:
Ceq = C1 + C2 + C3 + ……… + Cn
Where Ceq is the equivalent capacitance and C1, C2, C3, and Cn are the capacitance values of individual capacitors.
Using the formula above, we can obtain the equivalent capacitance of the capacitors connected in parallel as follows:
Ceq = 8.0 μF + 11 μF + 14 μF= 33 μF
Therefore, the equivalent capacitance of the capacitors connected in parallel is 33 μF.
Summing all of the individual capacitances in a circuit based on the relationships between these capacitors yields the equivalent capacitance, which is the sum of all of the capacitance values. Condensers, in particular, can be in series or parallel.
The idea of equivalent capacitance is used to show how one capacitor can replace multiple capacitors in a circuit. Therefore, the voltage drop for both a circuit with multiple capacitors connected to it and another circuit with a single capacitor of equivalent capacitance will be the same.
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A wire of length 10 meters carrying a current of .6 amps to the left lies along the x-axis from (-5,0) to (5,0) meters. a) Find the Magnetic field created by this wire at (0,8) meters. b) Find the Magnetic field created by this wire at (10,0) meters. c) Find the Magnetic field created by this wire at (10,8) meters.
The magnetic field created by the 10m wire carrying a current of 6A to the left lies along the x-axis from (-5,0) to (5,0) meters at:
a) point (0,8) m is approximately 3.75 × 10⁻⁹ T,
b) point (10,0) m is approximately 3 × 10⁻⁹ T and
c) point (10,8) m is approximately 2.68 × 10⁻⁹ T.
To find the magnetic field created by the wire at the given points, we can use the formula for the magnetic field produced by a straight current-carrying wire.
The formula is given by:
B = (μ₀ × I) / (2πr),
where
B is the magnetic field,
μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A),
I is the current, and
r is the distance from the wire.
a) At point (0,8) meters:The wire lies along the x-axis, and the point of interest is above the wire. The distance from the wire to the point is 8 meters. Substituting the values into the formula:
B = (4π × 10⁻⁷ T·m/A × 0.6 A) / (2π × 8 m),
B = (0.6 × 10⁻⁷ T·m) / (16 m),
B = 3.75 × 10⁻⁹ T.
Therefore, the magnetic field created by the wire at point (0,8) meters is approximately 3.75 × 10⁻⁹ T.
b) At point (10,0) meters:The wire lies along the x-axis, and the point of interest is to the right of the wire. The distance from the wire to the point is 10 meters. Substituting the values into the formula:
B = (4π × 10⁻⁷ T·m/A ×0.6 A) / (2π × 10 m),
B = (0.6 * 10⁻⁷ T·m) / (20 m),
B = 3 × 10⁻⁹ T.
Therefore, the magnetic field created by the wire at point (10,0) meters is approximately 3 × 10⁻⁹ T.
c) At point (10,8) meters:The wire lies along the x-axis, and the point of interest is above and to the right of the wire. The distance from the wire to the point is given by the diagonal distance of a right triangle with sides 8 meters and 10 meters. Using the Pythagorean theorem, we can find the distance:
r = √(8² + 10²) = √(64 + 100) = √164 = 4√41 meters.
Substituting the values into the formula:
B = (4π × 10⁻⁷ T·m/A × 0.6 A) / (2π × 4√41 m),
B = (0.6 × 10⁻⁷ T·m) / (8√41 m),
B ≈ 2.68 × 10⁻⁹ T.
Therefore, the magnetic field created by the wire at point (10,8) meters is approximately 2.68 × 10⁻⁹ Tesla.
Hence, the magnetic field created by the 10m wire carrying a current of 6A to the left lies along the x-axis from (-5,0) to (5,0) meters at a) point (0,8) meters is approximately 3.75 × 10⁻⁹ T, b) point (10,0) meters is approximately 3 × 10⁻⁹ T and c) point (10,8) meters is approximately 2.68 × 10⁻⁹ Tesla.
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Consider a container of nitrogen gas molecules at 900 K . Calculate.(b) the average speed.
The formula to calculate the average speed of gas particles is:Average speed of gas particles = √(8RT/πM) where R is the universal gas constant (8.31 J/Kmol), T is the temperature in Kelvin, and M is the molar mass of the gas.
Nitrogen gas molecules are present in a container at a temperature of 900K. The average speed of gas particles is to be calculated. We know that: Molar mass of nitrogen (N2) = 28 g/mol
R = 8.31 J/Kmol
T = 900 K
Now, we can substitute these values in the formula mentioned above.Average speed of gas particles = √(8RT/πM)
= √[(8 × 8.31 × 900)/(π × 28)]≈ 506.2 m/s
Therefore, the average speed of nitrogen gas molecules at a temperature of 900 K is approximately 506.2 m/s. The average speed of gas particles is the root mean square speed of the gas particles.
The formula for calculating the average speed of gas particles is:Average speed of gas particles = √(8RT/πM)
where R is the universal gas constant,
T is the temperature in Kelvin, and
M is the molar mass of the gas.
In this problem, we have nitrogen gas molecules present in a container at a temperature of 900K. The molar mass of nitrogen is 28 g/mol and the value of R is 8.31 J/Kmol. By substituting these values in the formula, we can calculate the average speed of nitrogen gas molecules which is approximately 506.2 m/s.
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An airline flight attendant rolls a suitcase through the airport lobby, as shown in the figure. If the magnitude of the force she exerts on the suitcase is 25.0 N, and she does +1.01×103] of work in moving the suitcase a distance of 53.0 m, at what angle θ above the horizontal (as shown in the figure above) is the force oriented with respect to the floor? 1 calorie =4.184 J
The force exerted by the flight attendant is oriented at an angle of approximately 41.14° above the horizontal with respect to the floor.
The formula for work is given as:
work = force * distance * cos(θ)
We are given the force exerted by the flight attendant on the suitcase as 25.0 N and the distance moved as 53.0 m. We also know that the work done is 1.01×10³ J.
Substituting these values into the formula, we get:
1.01×10³ J = 25.0 N * 53.0 m * cos(θ)
To find the angle θ, we rearrange the equation:
cos(θ) = 1.01×10³ J / (25.0 N * 53.0 m)
cos(θ) = 1.01×10³ J / (1325 N·m)
Using the conversion 1 calorie = 4.184 J, we can convert the units:
cos(θ) = (1.01×10³ J) / (1325 N·m) * (1 cal / 4.184 J)
cos(θ) = (1.01×10³ / 1325) cal / 4.184 N·m
cos(θ) ≈ 0.76015 cal / N·m
Now, to find the angle θ, we take the inverse cosine (cos⁻¹) of both sides:
θ ≈ cos⁻¹(0.76015 cal / N·m)
Using the calculator, we find:
θ ≈ 41.14°
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(a) If it takes 2.45 min to fill a 21.0 L bucket with water flowing from a garden hose of diameter 3.30 cm, determine the speed at which water is traveling through the hose. m/s (b) If a nozzle with a diameter three-fifths the diameter of the hose is attached to the hose, determine the speed of the water leaving the nozzle. m/s
The speed at which water is traveling through the hose is 0.1664 m/s. The speed of the water leaving the nozzle is 0.1569 m/s.
(a)If it takes 2.45 min to fill a 21.0L bucket with water flowing from a garden hose of diameter 3.30 cm, determine the speed at which water is traveling through the hose. m/s
Given that time taken to fill the 21.0 L bucket = 2.45 min Volume of water flowed through the hose = Volume of water filled in the bucket= 21.0 L = 21.0 × 10⁻³ m³Time taken = 2.45 × 60 = 147s Diameter of the hose, d₁ = 3.30 cm = 3.30 × 10⁻² m
The formula used to calculate speed of the water through the hose = Flow rate / Area of cross-section of the hose. Flow rate of water = Volume of water / Time taken.= 21.0 × 10⁻³ / 147= 1.428 × 10⁻⁴ m³/s Area of cross-section of the hose = 1/4 π d₁²= 1/4 × π × (3.30 × 10⁻²)²= 8.55 × 10⁻⁴ m²
Now, speed of water flowing through the hose is given byv = Q / A where Q = flow rate = 1.428 × 10⁻⁴ m³/sA = area of cross-section of the hose = 8.55 × 10⁻⁴ m²Substituting the values in the formula: v = 1.428 × 10⁻⁴ / 8.55 × 10⁻⁴= 0.1664 m/s Therefore, the speed at which water is traveling through the hose is 0.1664 m/s.
(b) If a nozzle with a diameter three-fifths the diameter of the hose is attached to the hose, determine the speed of the water leaving the nozzle. m/s Given that the diameter of the nozzle = 3/5 (3.30 × 10⁻²) m = 0.0198 m
The area of cross-section of the nozzle = 1/4 π d²= 1/4 × π × (0.0198)²= 3.090 × 10⁻⁵ m²The volume of water discharged by the nozzle is the same as that discharged by the hose.
V₁ = V₂V₂ = π r² h where r = radius of the nozzleh = height of water column V₂ = π (0.0099)² h = π (0.0099)² (21 × 10⁻³) = 6.11 × 10⁻⁵ m³The time taken to fill the bucket is the same as the time taken to discharge the volume of water from the nozzle. V₂ = Q t where Q = flow rate of water from the nozzle.
Substituting the value of V₂= Q × t = (6.11 × 10⁻⁵) / 2.45 × 60Q = 4.84 × 10⁻⁶ m³/s The speed of the water leaving the nozzle is given byv = Q / A where Q = flow rate = 4.84 × 10⁻⁶ m³/sA = area of cross-section of the nozzle = 3.090 × 10⁻⁵ m²Substituting the values in the formula: v = 4.84 × 10⁻⁶ / 3.090 × 10⁻⁵= 0.1569 m/s Therefore, the speed of the water leaving the nozzle is 0.1569 m/s.
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If the charge is -33_ μC, the speed is 1500_m/s, the strength of the magnetic field is 1_T, and the angle is 150∘, then find the force (magnitude and direction) on the charge. 2. magnitude A. 0.01548_N D. 0.02896_N B. 0.02475 N E. 0.03607 N C. 0.02817_N F. 0.02976_N 3. direction A. Left B. Into the paper C. Right D. Out of the paper
Given the charge, speed, magnetic field strength, and angle, we can calculate the force on the charge using the equation F = q * v * B * sin(θ). The magnitude of the force is 0.02896 N, and the direction is out of the paper.
The equation to calculate the force (F) on a moving charge in a magnetic field is given by F = q * v * B * sin(θ), where q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
Given:
Charge (q) = -33 μC = -33 × 10^-6 C
Speed (v) = 1500 m/s
Magnetic field strength (B) = 1 T
Angle (θ) = 150°
First, we need to convert the charge from microcoulombs to coulombs:
q = -33 × 10^-6 C
Now we can substitute the given values into the equation to calculate the force:
F = q * v * B * sin(θ)
= (-33 × 10^-6 C) * (1500 m/s) * (1 T) * sin(150°)
≈ 0.02896 N
Therefore, the magnitude of the force on the charge is approximately 0.02896 N.
To determine the direction of the force, we need to consider the right-hand rule. When the charge moves with a velocity (v) at an angle of 150° to the magnetic field (B) pointing into the paper, the force will be directed out of the paper.
Hence, the direction of the force on the charge is out of the paper.
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A hawk is flying from the sky vertically toward a rabbit on the ground with a velocity of 30m/s. It emits a screech at 3300 Hz to scare the rabbit. What is the frequency heard by the rabbit? Assume the screeching sound is reflected from the ground back towards the hawk, what is the frequency of the screech heard by the hawk? You may assume the velocity of the sound in air is 340m/s.
"The frequency heard by the rabbit is approximately 3064.86 Hz & the frequency heard by the hawk is approximately 3925.81 Hz."
To determine the frequency heard by the rabbit and the frequency heard by the hawk, we need to consider the Doppler effect. The Doppler effect describes the change in frequency of a wave as observed by an observer moving relative to the source of the wave.
Let's calculate the frequency heard by the rabbit first:
From question:
Velocity of the hawk (source): v_source = 30 m/s (moving vertically downwards)
Velocity of sound in air: v_sound = 340 m/s
The formula for the frequency heard by the observer (rabbit) is given by:
f_observed = (v_sound + v_observer) / (v_sound + v_source) * f_source
In this case, the observer (rabbit) is stationary on the ground, so the velocity of the observer is zero (v_observer = 0). Plugging in the values:
f_observed = (340 m/s + 0 m/s) / (340 m/s + 30 m/s) * 3300 Hz
f_observed = (340 m/s) / (370 m/s) * 3300 Hz
f_observed = 3064.86 Hz
Therefore, the frequency heard by the rabbit is approximately 3064.86 Hz.
Now let's calculate the frequency heard by the hawk:
In this case, the hawk is the observer, and the source of the sound is the reflection of its own screech from the ground.
From question:
Velocity of the hawk (observer): v_observer = 30 m/s (moving vertically downwards)
The velocity of sound in air: v_sound = 340 m/s
Using the same formula as before:
f_observed = (v_sound + v_observer) / (v_sound + v_source) * f_source
f_observed = (340 m/s + 30 m/s) / (340 m/s - 30 m/s) * 3300 Hz
f_observed = (370 m/s) / (310 m/s) * 3300 Hz
f_observed = 3925.81 Hz
Therefore, the frequency heard by the hawk is approximately 3925.81 Hz.
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A tangential force of 89789.9 N is applied to a 6.2mm copper cube as shown below, find the shear strain given that the shear modulus of brass is 4.2 X 1070 N/m?? A Shear strain rad
The shear strain of the copper cube is approximately 0.02144 radians.
To find the shear strain of the copper cube, we can use the equation:
Shear strain = Shear stress / Shear modulus
The applied tangential force is 89789.9 N and the shear modulus of brass (assuming it was mistakenly mentioned as copper) is 4.2 x 10^7 N/m², we need to convert the dimensions of the cube to obtain the shear stress.
The shear stress can be calculated using the formula:
Shear stress = Force / Area
The area of the cube's face can be determined by squaring the length of one side, which is given as 6.2 mm or 0.0062 m.
Now, let's calculate the shear stress:
Area = (0.0062 m)² = 3.844 x 10^-5 m²
Shear stress = 89789.9 N / 3.844 x 10^-5 m²
Next, we can calculate the shear strain:
Shear strain = Shear stress / Shear modulus
Shear strain = (89789.9 N / 3.844 x 10^-5 m²) / (4.2 x 10^7 N/m²)
Evaluating the expression, we find that the shear strain is approximately 0.02144 rad.
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Four objects are located on the Y axis: the 2.0 Kg object is 3.0 m from the origin; the 3.0 kg one is 2.5 m from the origin; the 2.5 kg one is at the origin; and the 4.0 Kg is located -0.50 m from the origin. Where is the center of mass of these objects?
The answer is, "The center of mass of these objects is located 0.83 meters from the origin."
To find out the center of mass of a set of objects, the following formula can be used:
[tex]\frac{\sum m_ix_i}{\sum m_i}[/tex]
where $m_i$ is the mass of the object, and $x_i$ is its distance from a reference point.
The values can be substituted into the formula to get the center of mass. So let's compute the center of mass of these objects:
[tex]\frac{(2.0\text{ Kg})(3.0\text{ m}) + (3.0\text{ Kg})(2.5\text{ m}) + (2.5\text{ Kg})(0.0\text{ m}) + (4.0\text{ Kg})(-0.50\text{ m})}{2.0\text{ Kg} + 3.0\text{ Kg} + 2.5\text{ Kg} + 4.0\text{ Kg}}\\=\frac{6.0\text{ Kg m}+7.5\text{ Kg m}-2.0\text{ Kg m}-2.0\text{ Kg m}}{11.5\text{ Kg}}\\=\frac{9.5\text{ Kg m}}{11.5\text{ Kg}}\\=0.83\text{ m}[/tex]
Therefore, the center of mass of the four objects is located at 0.83 meters from the origin.
The answer is, "The center of mass of these objects is located 0.83 meters from the origin."
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A voltage source E-5V is connected in series to a capacitance of 1 x 10 farad and a resistance of 4 ohms. What is the appropriate equation to model the behavior of the charge. Q. 100+ 4Q = 5 4 + 10 "Q-5 540 +10°Q = 4 de 04+109Q = 5 dr
The appropriate equation to model the behavior of the charge is Q - 5 + 10⁹Q = 4.
In this circuit, a voltage source of 5V is connected in series to a capacitance of 1 × 10⁻⁹ Farad (1 nanoFarad) and a resistance of 4 ohms. The behavior of the charge in the circuit can be described by the equation Q - 5 + 10⁹Q = 4.
Let's break down the equation:
Q represents the charge in Coulombs on the capacitor.
The first term, Q, accounts for the charge stored on the capacitor.
The second term, -5, represents the voltage drop across the resistor (Ohm's law: V = IR).
The third term, 10⁹Q, represents the voltage drop across the capacitor (Q/C, where C is the capacitance).
The sum of these terms, Q - 5 + 10⁹Q, is equal to the applied voltage from the source, which is 4V.
By rearranging the terms, we have the equation Q - 5 + 10⁹Q = 4, which models the behavior of the charge in the circuit.
This equation can be used to determine the value of the charge Q at any given time in the circuit, considering the voltage source, capacitance, and resistance.
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A vertical pole of height h = 1.000 m is standing on the ground of an empty swimming pool. It casts a shadow of length d1 = 0.577 m on the floor of the pool. a) The pool is now filled with water up to a height of 1 m. How long (dz) is the shadow on the bottom of the pool? The index of refraction of water is 1.33. b) How fast does the light travel in water? c) If light has a wavelength of 800 nm in air, what is the wavelength in water?
a) The pool is now filled with water up to a height of 1 m. 0.2885 meters long (dz) is the shadow on the bottom of the pool.
b) The light travel in water at 2.26 x 10^8 meters per second.
c) If light has a wavelength of 800 nm in air, the wavelength in water is 601.5 nanometers.
To solve this problem, we can use the concept of similar triangles and Snell's law.
a) Finding the length of the shadow (dz) on the bottom of the pool when it is filled with water:
Let's assume the height of the shadow on the bottom of the pool is dz.
According to similar triangles, we can set up the following proportion:
dz / h = d1 / (h + 1)
Substituting the given values:
dz / 1.000 = 0.577 / (1.000 + 1)
dz = (0.577 * 1.000) / 2.000
dz = 0.2885 m
Therefore, the length of the shadow on the bottom of the pool when it is filled with water is approximately 0.2885 meters.
b) Calculating the speed of light in water:
The speed of light in a medium can be determined using the formula:
v = c / n
Where:
v = Speed of light in the medium
c = Speed of light in vacuum (approximately 3.00 x 10^8 m/s)
n = Refractive index of the medium
Substituting the values:
v = (3.00 x 10^8 m/s) / 1.33
v ≈ 2.26 x 10^8 m/s
Therefore, the speed of light in water is approximately 2.26 x 10^8 meters per second.
c) Calculating the wavelength of light in water:
The wavelength of light in a medium can be calculated using the formula:
λ = λ0 / n
Where: k
λ = Wavelength of light in the medium
λ0 = Wavelength of light in vacuum or air
n = Refractive index of the medium
Substituting the given values:
λ = 800 nm / 1.33
λ ≈ 601.5 nm
Therefore, the wavelength of light in water is approximately 601.5 nanometers.
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A sample of methane gas undergoes a change which causes it's pressure to decrease to 1/2 of it's original pressure, at the same time the volume increases by a factor of 4 . If the original temperature was 210C, what was the final temperature? If 16.4 moles of gas added to a system cause it's pressure to increase from 0.5×10 5Pa to 1.6 atm at constant volume and temperature. How many moles of gas was in the system in the end?
The pressure of methane gas decreases to half its original pressure while its volume increases by a factor of 4. The final temperature is approximately 60.39 K. There were 16.4 moles of gas in the system at the end.
To solve these problems, we can use the ideal gas law, which states:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
1. Sample of Methane Gas:
According to the problem, the pressure decreases to 1/2 of its original value, and the volume increases by a factor of 4. The temperature is also given.
Let's assume the original pressure is P1, the final pressure is P2, the original volume is V1, the final volume is V2, the original temperature is T1, and the final temperature is T2.
We have the following information:
P2 = 1/2 * P1
V2 = 4 * V1
T1 = 210°C
First, we need to convert the temperature to Kelvin since the ideal gas law requires temperature in Kelvin. To convert Celsius to Kelvin, we add 273.15:
T1(K) = T1(°C) + 273.15
T1(K) = 210 + 273.15 = 483.15 K
Now, we can use the ideal gas law to relate the initial and final states of the gas:
(P1 * V1) / T1 = (P2 * V2) / T2
Substituting the given values:
(P1 * V1) / 483.15 = (1/2 * P1 * 4 * V1) / T2
Simplifying the equation:
4P1V1 = 483.15 * P1 * V1 / (2 * T2)
Canceling out P1V1:
4 = 483.15 / (2 * T2)
Multiplying both sides by 2 * T2:
8 * T2 = 483.15
Dividing both sides by 8:
T2 = 60.39375 K
Therefore, the final temperature is approximately 60.39 K.
2. Adding Moles of Gas: In this problem, the pressure increases from 0.5 × 10⁵ Pa to 1.6 atm at constant volume and temperature. The number of moles of gas added is given as 16.4 moles.
Let's assume the initial number of moles is n1, and the final number of moles is n2. We know that the pressure and temperature remain constant, so we can use the ideal gas law to relate the initial and final number of moles:
(P1 * V) / (n1 * R * T) = (P2 * V) / (n2 * R * T)
Canceling out V, P1, P2, and R * T:
1 / (n1 * R) = 1 / (n2 * R)
Now, we can solve for n2:
1 / n1 = 1 / n2
n2 = n1
Since the initial number of moles is n1 = 16.4 moles, the final number of moles is also 16.4 moles. Therefore, there were 16.4 moles of gas in the system at the end.
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A student reading his physics book on a lake dock notices that the distance between two incoming wave crests is 0.75 m, and he then measures the time of arrival between the crests to be 1.6 s. Determine the (i) frequency (ii) speed of the waves.
The frequency of the wave is 0.625 Hz. The speed of the wave is 0.469 m/s.
Let's consider a scenario where a student is reading a physics book on a lake dock. The student observes that there is a distance of 0.75 meters between two consecutive wave crests. Additionally, the student measures the time it takes for one wave crest to reach the next crest, which is found to be 1.6 seconds. Now, we can proceed to determine the (i) frequency and (ii) speed of the waves.
(i) Frequency:
We know that frequency is the number of wave cycles that pass a point in one second. This is denoted by f and has units of hertz (Hz).We can use the formula:
frequency = 1 / time period
Given that the time taken for one wave crest to reach the next wave crest is measured to be 1.6 seconds,
frequency = 1 / time period= 1 / 1.6 s= 0.625 Hz
Therefore, the frequency of the wave is 0.625 Hz.
(ii) Speed:We can use the formula for wave speed:
v = frequency × wavelength
Given the distance between two incoming wave crests is 0.75 m, we can get the wavelength by:
wavelength = distance between two incoming wave crests= 0.75 m
Given the frequency is 0.625 Hz,v = frequency × wavelength= 0.625 × 0.75= 0.469 m/s
Therefore, the speed of the wave is 0.469 m/s.
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A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.40 cm wide and 11.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm
thick.
What is the maximum charge that can be stored in this capacitor?
We can calculate the capacitance of the capacitor:
C = ε₀(A/d) = (8.85 x [tex]10^{-12}[/tex] F/m) × (0.374 m² / 0.0000225 m)
≈ 1.467 x [tex]10^{-9}[/tex] F.
To find the maximum charge that can be stored in the capacitor, we need to use the formula for the capacitance of a parallel-plate capacitor:
C = ε₀(A/d)
Where:
- C is the capacitance.
- ε₀ is the vacuum permittivity, approximately equal to 8.85 x 10^(-12) F/m.
- A is the area of the plates.
- d is the separation between the plates.
Given:
- The width of each aluminum-foil sheet is 3.40 cm = 0.034 m.
- The length of each aluminum-foil sheet is 11.0 m.
- The mica strip has the same width and length.
- The thickness of the mica strip is 0.0225 mm = 0.0000225 m.
First, let's calculate the area of each plate:
A = width × length
= 0.034 m × 11.0 m
= 0.374 m²
Determine the effective separation between the plates.
d = thickness of mica + thickness of air gap
= 0.0000225 m + 0 (since air gap is negligible)
= 0.0000225 m
Now, we can calculate the capacitance of the capacitor:
C = ε₀(A/d) = (8.85 x [tex]10^{-12}[/tex] F/m) × (0.374 m² / 0.0000225 m)
≈ 1.467 x[tex]10^{-9}[/tex] F
Finally, the maximum charge that can be stored in the capacitor is given by the equation:
Q = C × V
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The maximum charge that can be stored in this capacitor is 6.46 x 10^-5 C.
The maximum charge that can be stored in the parallel-plate capacitor is 6.46 x 10^-5 C. Capacitance is the ability of an object to store an electric charge, and it is determined by the size, shape, and distance between the plates. Here, a parallel-plate capacitor is made from two aluminum-foil sheets, each 3.40 cm wide and 11.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick.
The capacitance of a parallel plate capacitor is given by;C=εA/d,where ε is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.ε = 8.85 x 10^-12 F/m is the permittivity of free space A = (3.40 x 10^-2 m) x (11.0 m) = 0.374 m^2 is the area of each plated = 0.0225 x 10^-3 m is the distance between the plates
Therefore, the capacitance is;C=εA/d = 8.85 x 10^-12 x 0.374/0.0225 x 10^-3 = 1.47 x 10^-8 FThe maximum charge that can be stored in a capacitor is given by;Q=CV, where Q is the maximum charge, C is the capacitance, and V is the voltage applied across the capacitor.
To find the maximum charge, we can use the voltage equation,V=Ed/d = εE/d,where E is the electric field between the plates and d is the distance between the plates. Since the electric field is uniform, we have;E=V/d = εV/d^2Substituting the expression for the electric field into the capacitance equation, we have;C=εA/d = εA/V/ESimplifying for the voltage, we have;V=Q/CSubstituting the expression for the electric field into the voltage equation, we have;Q = CV = εAV/dThe maximum charge that can be stored in this capacitor is thus;Q = εAV/d = 8.85 x 10^-12 x 0.374/0.0225 x 10^-3 = 6.46 x 10^-5 C
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