The telescope can resolve objects with an angular size greater than or equal to 1.21 arcseconds.
The resolving power of a telescope determines its ability to distinguish fine details in an observed object. It is determined by the diameter of the objective lens or mirror and the wavelength of the light being observed. The formula for resolving power is given by:
R = 1.22 * (λ / D)
Where R is the resolving power, λ is the wavelength of light, and D is the diameter of the objective lens or mirror.
In this case, the diameter of the objective lens is given as 1.00 m, and the wavelength of green light is 550 nm (or 550 x 10^-9 m). Plugging in these values into the formula, we can calculate the resolving power:
R = 1.22 * (550 x 10^-9 m / 1.00 m)
R ≈ 1.21 x 10^-3 radians
To convert the resolving power to angular size, we can use the fact that there are approximately 206,265 arcseconds in a radian:
Angular size = R * (206,265 arcseconds/radian)
Angular size ≈ 1.21 x 10^-3 radians * 206,265 arcseconds/radian
The result is approximately 1.21 arcseconds. Therefore, the telescope can resolve objects with an angular size greater than or equal to 1.21 arcseconds.
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52. Find the approximate gravitational red shift in 500-nm light emitted by a white dwarf star whose mass is that of the sun but whose radius is that of the earth, 6.4 X 105 m.
The approximate gravitational red shift in 500-nm light emitted by a white dwarf star whose mass is that of the sun but whose radius is that of the earth, 6.4 x 105 m is 5.96 x 10-6. The gravitational redshift is defined as the decrease in frequency and energy of a photon as it moves from a higher gravitational potential to a lower one. Gravitational redshift happens because of the effect of gravity on light.
Explanation:
The gravitational red shift is given by
Δλ/λ = GM/(Rc²)
where
Δλ/λ = fractional shift of the wavelength of light.
G = gravitational constant (6.67 × 10-11 Nm²/kg²)
M = mass of the object (1 M☉ = 1.99 × 10³⁰ kg)
R = radius of the object (earth radius, 6.4 × 10⁶ m)
c = speed of light (3 × 10⁸ m/s)
Substitute the values in the above formula
Δλ/λ = (6.67 × 10-11 Nm²/kg²) × (1.99 × 1030 kg) / [(6.4 × 106 m) × (3 × 108 m/s)²]
Δλ/λ = 5.96 × 10-6
Therefore, the approximate gravitational red shift in 500-nm light emitted by a white dwarf star whose mass is that of the sun but whose radius is that of the earth, 6.4 × 105 m is 5.96 × 10-6.
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6. The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz?. Determine the electric field in this region at the coordinate (3,4,5).
The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz.The electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.
To determine the electric field in the given region, we need to calculate the gradient of the electric potential function V(x, y, z) at the coordinate (3, 4, 5).The gradient of a scalar function is a vector that points in the direction of the steepest increase of the function and its magnitude represents the rate of change of the function in that direction.
The electric potential function is given as V(x, y, z) = x^2 + xy^2 + 2yz.
To find the gradient, we need to calculate the partial derivatives of V with respect to each coordinate (x, y, z):
∂V/∂x = 2x + y^2
∂V/∂y = 2xy
∂V/∂z = 2y
Now, we can evaluate these partial derivatives at the coordinate (3, 4, 5):
∂V/∂x = 2(3) + (4)^2 = 6 + 16 = 22
∂V/∂y = 2(3)(4) = 24
∂V/∂z = 2(4) = 8
Therefore, the electric field at the coordinate (3, 4, 5) is given by the vector E = -(∂V/∂x)i - (∂V/∂y)j - (∂V/∂z)k:
E = -22i - 24j - 8k
So, the electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.
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TWO-Dimensiona Solve for Distance, Time, and Constant Velocity: 1) A police officer in a police car finds that a vehicle is travelling beyond the speed limit in a low-velocity zone with a constant speed of 24 m/s. As soon as the vehicle passes the police car, the police officer begins pursuing the vehicle with a constant acceleration of 6 m/s2 until the police office catches up with and stops the speeding vehicle. (NOTE: here the distance covered, and the time elapsed, is the same for both the POLICE CAR and the SPEEDING VEHICLE, from the time the police car begins pursuing the vehicle to the time the police car catches up and stops the vehicle). A) What is the time taken by the police car to catch up with and stop the speeding vehicle?
Given that a police officer in a police car finds that a vehicle is travelling beyond the speed limit in a low-velocity zone with a constant speed of 24 m/s. As soon as the vehicle passes the police car, the police officer begins pursuing the vehicle with a constant acceleration of 6 m/s² until the police office catches up with and stops the speeding vehicle. Here, the distance covered and the time elapsed are the same for both the POLICE CAR and the SPEEDING VEHICLE, from the time the police car begins pursuing the vehicle to the time the police car catches up and stops the vehicle.
The time taken by the police car to catch up with and stop the speeding vehicle is 4 seconds.
We need to find the time taken by the police car to catch up with and stop the speeding vehicle.
Solution:
Let the time taken to catch up with and stop the vehicle be t.
So, the distance covered by the police car during the time t = distance covered by the speeding vehicle during the time Distance = speed × time.
Distance covered by the speeding vehicle during the time t is 24t.
Distance covered by the police car during the time t is 1/2 × 6t², since it starts from rest and its acceleration is 6 m/s².
We know that both distances are the same.
Therefore, 24t = 1/2 × 6t²
⇒ 4t = t²
⇒ t = 4 s.
Therefore, the time taken by the police car to catch up with and stop the speeding vehicle is 4 seconds.
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"A ray of light strikes a surface at ninety degrees, that is, it
is parallel to the normal. The angle of refraction is
A. one hundred and eighty degrees, 180°
B. ninety degrees, 90°.
C. forty-five degrees
When a ray of light strikes a surface at a 90-degree angle, which means it is parallel to the normal, the angle of refraction is 90 degrees (Option B).
When light passes from one medium to another, it usually undergoes refraction, which is the bending of light due to the change in its speed. The angle of refraction is determined by Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the two media.
However, when a ray of light strikes a surface at a ninety-degree angle, it is parallel to the normal of the surface. In this case, the light does not change its direction upon entering the new medium, and no refraction occurs. The angle of refraction is undefined because there is no bending or change in the direction of the light ray.
Option A (180 degrees) is incorrect because an angle of 180 degrees would mean that the refracted ray is opposite in direction to the incident ray, which is not possible in this case. Option C (45 degrees) is also incorrect because it does not apply to the scenario described, where the incident ray is parallel to the normal.
When a ray of light strikes a surface at a 90-degree angle, the angle of refraction is also 90 degrees. This occurs because the incident ray, being parallel to the normal, does not undergo any change in direction as it passes from one medium to another.
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7. A radio station broadcasts its radio signals at 92.6 MHz. Find the wavelength if the waves travel at 3.00 x 108 m/s.
The problem involves a radio station broadcasting at a frequency of 92.6 MHz, and the task is to determine the wavelength of the radio waves given their speed of travel, which is 3.00 x 10^8 m/s.
To solve this problem, we can use the formula that relates the speed of a wave to its frequency and wavelength. The key parameters involved are frequency, wavelength, and speed.
The formula is: speed = frequency * wavelength. Rearranging the formula, we get: wavelength = speed / frequency. By substituting the given values of the speed (3.00 x 10^8 m/s) and the frequency (92.6 MHz, which is equivalent to 92.6 x 10^6 Hz), we can calculate the wavelength of the radio waves.
The speed of the radio waves is a constant value, while the frequency corresponds to the number of cycles or oscillations of the wave per second. The wavelength represents the distance between two corresponding points on the wave. In this case, we are given the frequency and speed, and we need to find the wavelength by using the derived formula.
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6. Design an absorption packed tower that is used to reduce NH; in air from a concentration of 0.10 kg'm' to a concentration of 0.0005 kg/ml Given: Column diameter - 3.00 m Operating temperature - 20.0°C Air density at 20.0°C -1.205 kg/m Operating pressure 101.325 kPa For 15 kg NH, per 100 kg H:0 measured partial pressure of NH3 – 15.199 kPa Q-01-10.0 kg's H-0.438 m H=0.250 m Incoming liquid is water free of NH3(x2) = 0 GMW of NH) - 17.030 GMW of Air - 28.970 GMW of H20 - 18.015 Determine 6.1 Mole fraction of pollutant in the gas phase at inlet of tower (y) 6.2 Equilibrium mole fraction of pollutant in the liquid phase (3) 6.3 Slope of equilibrium curve (m) 6.4 Absorption factor (A) 6.5 Height of an absorption packed tower (2)
The objective is to design an absorption packed tower to reduce NH3 concentration in air, and the parameters to be determined are the mole fraction of pollutant in the gas phase at the tower inlet (y), the equilibrium mole fraction of pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the tower.
What is the objective of the given problem and what parameters are to be determined for designing the absorption packed tower for NH3 removal?The given problem involves the design of an absorption packed tower for removing NH3 from air. The tower should reduce the NH3 concentration from 0.10 kg/m3 to 0.0005 kg/m3.
The operating conditions include a column diameter of 3.00 m, operating temperature of 20.0°C, air density at 20.0°C of 1.205 kg/m3, and operating pressure of 101.325 kPa. The relevant data includes the measured partial pressure of NH3, the flow rate of H2O, and the molecular weights of NH3, air, and H2O.
The objectives are to determine the mole fraction of the pollutant in the gas phase at the inlet of the tower (y), the equilibrium mole fraction of the pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the absorption packed tower.
These parameters will help in designing an effective tower for NH3 removal.
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A police car with a mass of 1800 kg is headed west at 60 km/h when it has an inelastic collision with a southbound 4500 kg ambulance. The wreckage ended up travelling at 41 km/h at 65° south of west.
What was the initial velocity and direction of the ambulance? Show your vector diagram. Be sure to label your diagram and indicate direction
Is the collision above an elastic or inelastic collision? How do you know?
The collision is an inelastic collision.This can be known because in an inelastic collision, the objects stick together and move with a common velocity after the collision.
The initial velocity and direction of the ambulance:The initial velocity and direction of the ambulance can be calculated using the conservation of momentum principle which states that the total momentum before a collision is equal to the total momentum after the collision.
A police car of 1800 kg is heading west at 60 km/h and a southbound ambulance of 4500 kg has an unknown initial velocity.
Let the initial velocity of the ambulance be u m/s at angle θ with respect to the horizontal such that:u cos θ is the horizontal component of the initial velocity.u sin θ is the vertical component of the initial velocity.
Momentum before collision = Momentum after collision
Thus:1800(60) + 0 = 1800v + 4500v cos 65° + 4500v sin 65°1800v = 108000 – 34891.924v = 57.77 km/h
Let the angle the wreckage makes with the west direction be θ2. Using vector addition,The horizontal component of the wreckage velocity = v cos 65°
The vertical component of the wreckage velocity = v sin 65°
The magnitude of the wreckage velocity is 41 km/h.
Then:tanθ2 = (v sin 65°) / (v cos 65°)θ2 = 50.59° south of west
Thus the initial velocity and direction of the ambulance are 57.77 km/h at 50.59° south of west.
Therefore the collision above is an inelastic collision. This can be known because in an inelastic collision, the objects stick together and move with a common velocity after the collision. The wreckage continued to move together as a single entity after the collision.
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In a purely inductive AC circuit as shown in the figure, AV, = 100 V. max AVmax sin wt L 000 (a) The maximum current is 5.00 A at 40.0 Hz. Calculate the inductance L. H (b) At what angular frequency w is the maximum current 1.50 A? rad/s
(a) The inductance (L) in the purely inductive AC circuit is approximately 79.6 mH, and (b) the angular frequency (ω) at which the maximum current is 1.50 A is approximately 838.93 rad/s.
(a) To calculate the inductance (L) in a purely inductive AC circuit, we can use the formula relating the maximum current (Imax), angular frequency (ω), and inductance (L).
The formula is Imax = (Vmax / ωL), where Vmax is the maximum voltage. Rearranging the formula, we have L = Vmax / (Imax ω). Plugging in the given values of Imax = 5.00 A and ω = 2πf = 2π × 40.0 Hz, and Vmax = 100 V, we can calculate L as L = 100 V / (5.00 A × 2π × 40.0 Hz) ≈ 0.0796 H or 79.6 mH.
(b) To find the angular frequency (ω) at which the maximum current (Imax) is 1.50 A, we can rearrange the formula used in part (a) as ω = Vmax / (Imax L).
Plugging in the given values of Imax = 1.50 A, Vmax = 100 V, and L = 79.6 mH (0.0796 H), we can calculate ω as ω = 100 V / (1.50 A × 0.0796 H) ≈ 838.93 rad/s.
In summary, (a) the inductance (L) in the purely inductive AC circuit is approximately 79.6 mH, and (b) the angular frequency (ω) at which the maximum current is 1.50 A is approximately 838.93 rad/s.
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Achild on a merry-go-round takes \( 1.9 \). 8 to go around once. What is his angular displacement during a \( 1.0 \) s tirno interval? Exprese your answer in radlans.
The child's angular displacement during the 1.0-second time interval is 3.30 radians. Angular displacement is the change in the position or orientation of an object with respect to a reference point or axis.
To find the angular displacement of the child during a 1.0-second time interval, we can use the formula:
θ = ω * t
Where: θ is the angular displacement (unknown), ω is the angular velocity (in radians per second), t is the time interval (1.0 s)
Given that the child takes 1.9 seconds to go around once, we can determine the angular velocity as:
ω = (2π radians) / (1.9 s)
Substituting the values into the formula:
θ = [(2π radians) / (1.9 s)] * (1.0 s),
θ = 2π/1.9 radians
θ = 3.30 radians
Therefore, the child's angular displacement during the 1.0-second time interval is approximately 3.30 radians.
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Six
moles of an ideal gas, initially at 178°C, are compressed to one
third of their initial volume at a constant pressure of 2.00 atm.
What is the final temperature of the gas?
Seis moles de un gas ideal, inicialmente a 178 °C, se comprimen a un tercio de su volumen inicial a presión constante de 2.00 atm. ¿Cuál es la temperatura final del gas? Select one: a.-123 °C b.-
The final temperature of the gas after compression is approximately 150.38°C.
To determine the final temperature of the gas after compression, using the combined gas law:
(P₁ ×V₁) / T₁= (P₂ × V₂) / T₂
Where:
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Given:
P₁ = 2.00 atm (constant pressure)
V₁ = Initial volume
T₁ = 178°C + 273.15
P₂ = 2.00 atm (constant pressure)
V₂ = (1/3) × V₁
T₂ = Final temperature
Substituting the values and solving for T₂
(2.00 atm × V₁) / (178°C + 273.15) = (2.00 atm × (1/3) × V₁) / T₂
V₁ / (178°C + 273.15) = (1/3) × V₁ / T₂
T₂ = (178°C + 273.15) × (1/3)
T₂ ≈ 150.38°C
Therefore, the final temperature of the gas after compression is approximately 150.38°C.
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Resistor in circuit is made of a length of 14awg iron wire. When
10 V is applied across the resistor wire of length 100m,
what is the reading on the ammeter? The thickness
of 14awg wire is 1.628mm.
The reading on the ammeter would be approximately 2.14 Amperes.
To calculate the reading on the ammeter, we need to determine the resistance of the 14 AWG iron wire. The resistance can be calculated using the formula
[tex]R = ρ * (L / A)[/tex]
where:
R is the resistance,
ρ is the resistivity of the material (in this case, iron),
L is the length of the wire, and
A is the cross-sectional area of the wire.
First, let's calculate the cross-sectional area of the 14 AWG wire. The diameter of the wire can be obtained from the wire gauge size. For 14 AWG, the diameter is approximately 1.628 mm.
The radius (r) can be calculated by dividing the diameter by 2:
r = 1.628 mm / 2 = 0.814 mm = 0.000814 m
The cross-sectional area (A) can be calculated using the formula:
[tex]R = ρ * (L / A)[/tex]
[tex]A = 3.14159 * (0.000814 m)^2 ≈ 2.07678 × 10^(-6) m^2[/tex]
Next, we need to find the resistivity of iron. The resistivity of iron (ρ) is approximately 9.71 × 10^(-8) Ω·m.
Now, we can calculate the resistance (R) using the formula mentioned earlier:
[tex]R = (9.71 × 10^(-8) Ω·m) * (100 m / 2.07678 × 10^(-6) m^2)[/tex]
[tex]R ≈ 4.675 Ω[/tex]
Therefore, with a 10 V potential difference across the 14 AWG iron wire resistor, the reading on the ammeter would be:
[tex]I = V / R[/tex]
[tex]I = 10 V / 4.675 Ω[/tex]
[tex]I ≈ 2.14 A[/tex]
So, the reading on the ammeter would be approximately 2.14 Amperes.
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Consider two strings tuned to the note A (440 Hz), mounted on guitars of the same size. The high-tension string has a diameter of 0.432 mm, and the low-tension string has a diameter of 0.381 mm. The strings are made of the same material, so they have the same density p. The strings can be thought of as long cylinders. What is the ratio of the high tension to the low tension?
The ratio of high tension to low tension is `1.22`.Hence, option D is correct.
Given data: Frequency of both the string,
`f = 440 Hz`
Diameter of high tension string, `d_1 = 0.432 mm
`Diameter of low tension string, `d_2 = 0.381 mm`
The density of both strings is the same.
Let the tension in high tension string and low tension string be `T_1` and `T_2` respectively.
Ratio of tension in both strings:
`T_1/T_2= [(π/4)d_1²p(f₁)²]/[(π/4)d_2²p(f₂)²]`
Here, `f₁ = f₂ = f =
440 Hz`.
So,
`T_1/T_2 = d_1²/d_2² = (0.432)²/(0.381)²
≈ 1.22`
Therefore, the ratio of high tension to low tension is `1.22`.
Hence, option D is correct.
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The ratio of the high-tension to the low-tension is 1.3616:1.Given Data: Diameter of high tension string: d₁ = 0.432 mm Diameter of low tension string:
d₂ = 0.381 mm
The strings are made of the same material, so they have the same density p.
Frequency of both the strings: f = 440 Hz Formula Used:
The tension (T) in a string is given by, T = μf²d²π² Where, μ is the linear density of the string (mass per unit length)d is the diameter of the string f is the frequency of vibration of the stringπ = 3.14 Calculation:
Let the tension in the high-tension string be T₁ and the tension in the low-tension string be T₂ We know that,μ = pA where, p is the density of the string
A = πd²/4 is the cross-sectional area of the string As the strings are made of the same material, they have the same density.
Therefore,μ₁ = μ₂
⇒ pA₁ = pA₂
⇒ A₁ = A₂d₁²
= d₂²
= (0.432 mm)²
= 0.186624 mm²
= A₁A₂
= (0.381 mm)²
= 0.144961 mm²
Therefore, A₁/A₂ = (0.432 mm)²/(0.381 mm)²
= 1.3616/1T₁ = μf²d₁²π²and,T₂ = μf²d₂²π²Dividing these two equations,
T₁/T₂ = μ₁f²d₁²π²/μ₂f²d₂²π²
⇒ T₁/T₂ = d₁²/d₂²
⇒ T₁/T₂ = (0.432 mm)²/(0.381 mm)²
⇒ T₁/T₂ = 1.3616/1
⇒ T₁/T₂ = 1.3616:1
Therefore, the ratio of the high-tension to the low-tension is 1.3616:1.
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An electron has a kinetic energy of 2.4 eV. What is the wavelength of the electron?
Enter your answer in nm to two decimal places but do not enter the units
The wavelength of an electron with a kinetic energy of 2.4 eV can be calculated using the de Broglie wavelength equation. The wavelength, expressed in nanometers (nm) to two decimal places, can be determined numerically.
The de Broglie wavelength equation relates the wavelength (λ) of a particle to its momentum (p). For an electron, the equation is given by:
λ = h / p
Where:
λ is the wavelength,
h is the Planck's constant (approximately 6.626 x 10^-34 J·s), and
p is the momentum.
The momentum of an electron can be calculated using its kinetic energy (KE) and mass (m) through the equation:
p = sqrt(2 * m * KE)
To find the wavelength, we first need to convert the kinetic energy from electron volts (eV) to joules (J) using the conversion factor: 1 eV = 1.602 x 10^-19 J. Then, we can calculate the momentum and substitute it into the de Broglie wavelength equation.
By plugging in the appropriate values and performing the calculations, we can find the wavelength of the electron in nanometers to two decimal places.
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Suppose that an electron trapped in a one-dimensional infinite well of width 307 pm is excited from its first excited state to the state with n = 9. (a) What energy must be transferred to the electron for this quantum jump? The electron then de-excites back to its ground state by emitting light. In the various possible ways it can do this, what are the (b) shortest, (c) second shortest, (d) longest, and (e) second longest wavelengths that can be emitted?
a) The energy transferred to the electron for the quantum jump from the first excited state to the state with n = 9 is 1.52 eV.
b) The shortest wavelength emitted when the electron de-excites back to its ground state is approximately 410 nm.
c) The second shortest wavelength emitted is approximately 821 nm.
d) The longest wavelength emitted is approximately 4100 nm.
e) The second longest wavelength emitted is approximately 8210 nm.
a) The energy transferred to the electron for the quantum jump can be calculated using the formula for the energy levels of a particle in an infinite well. The energy of the nth level is given by Eₙ = (n²h²)/(8mL²), where h is the Planck's constant, m is the mass of the electron, and L is the width of the well. By calculating the energy difference between the first excited state (n = 2) and the state with n = 9, we can determine the energy transferred, which is approximately 1.52 eV.
b), c), d), e) When the electron de-excites back to its ground state, it emits light with various wavelengths. The wavelength can be determined using the formula λ = 2L/n, where λ is the wavelength, L is the width of the well, and n is the quantum number of the state.
The shortest wavelength corresponds to the highest energy transition, which occurs when n = 2. Substituting the values, we find the shortest wavelength to be approximately 410 nm.
Similarly, we can calculate the wavelengths for the second shortest, longest, and second longest emitted light, which are approximately 821 nm, 4100 nm, and 8210 nm, respectively. These values correspond to the different possible transitions the electron can undergo during de-excitation.
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The energy required for the electron to transition from its first excited state to the state with n = 9 can be calculated using a formula. The shortest, second shortest, longest, and second longest wavelengths that can be emitted when the electron de-excites can also be calculated using a formula.
Explanation:(a) The energy required for the electron to transition from its first excited state to the state with n = 9 can be calculated using the formula:
E = ((n^2)π^2ħ^2) / (2mL^2)
where n is the quantum number, ħ is the reduced Planck's constant, m is the mass of the electron, and L is the width of the infinite well.
(b) The shortest wavelength that can be emitted corresponds to the transition from the excited state with n = 9 to the ground state with n = 1. This can be calculated using the formula:
λ = 2L / n
(c), (d), and (e) The second shortest, longest, and second longest wavelengths that can be emitted correspond to other possible transitions from the excited state with n = 9 to lower energy states. These can be calculated using the same formula.
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A 190,000 kg space probe is landing on an alien planet with a gravitational acceleration of 5.00. If its fuel is ejected from the rocket motor at 40,000 m/s what must the mass rate of change of the space ship (delta m)/( delta t ) be to achieve at upward acceleration of 2.50 m/s ^ 2 ?
A roller coaster cart of mass 114.0 kg is pushed against a launcher spring with spring constant 550.0 N/m compressing it by 11.0 m in the process. When the roller coaster is released from rest the spring pushes it along the track (assume no friction in cart bearings or axles and no rolling friction between wheels and rail). The roller coaster then encounters a series of curved inclines and declines and eventually comes to a horizontal section where it has a velocity 7.0 m/s. How far above or below (vertical displacement) the starting level is this second (flat) level? If lower include a negative sign with the magnitude.
The mass rate of change of the space ship is 190,000 kg/s and the required displacement is 8.88 m (upwards).
Question 1A The space probe lands on an alien planet with a gravitational acceleration of 5.00 m/s².
Now, the upward acceleration required is 2.50 m/s². Hence, the required acceleration can be calculated as:
∆v/∆t = a Where,
∆v = change in velocity = 40,000 m/s
a = acceleration = 2.50 m/s²
∆t can be calculated as:
∆t = ∆v/a
= 40,000/2.5
= 16,000 seconds
Therefore, the mass rate of change of the space ship is calculated as:
∆m/∆t = (F/a)
Where, F = force
= m × a
F = (190,000 kg) × (2.5 m/s²)
F = 475,000 N
∆m/∆t = (F/a)∆m/∆t
= (475,000 N) / (2.5 m/s²)
∆m/∆t = 190,000 kg/s
Question 2 Mass of the roller coaster, m = 114 kg
Spring constant, k = 550 N/m
Compression, x = 11.0 m
Initial velocity of the roller coaster, u = 0
Final velocity of the roller coaster, v = 7.0 m/s
At point A (Start)
Potential Energy + Kinetic Energy = Total Energy
[tex]1/2 kx^2+ 0 = 1/2 mv^2 + mgh[/tex]
[tex]0 + 0 = 1/2 \times 114 \times 7^2 + 114 \times g \times h[/tex]
[tex]1/2 \times 114 \times 7^2 + 0 = 114 \times 9.8 \times h[/tex]
h = 16.43 m
At point B (End)
Potential Energy + Kinetic Energy = Total Energy
[tex]0 + 1/2 \ mv^2 = 1/2 \ mv^2 + mgh[/tex]
[tex]0 + 1/2 \times 114 \times 7^2= 0 + 114 \times 9.8 \times h[/tex]
h = -7.55 m
So, the vertical displacement is 16.43 m - 7.55 m = 8.88 m (upwards)
Therefore, the required displacement is 8.88 m (upwards).
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The vertical displacement from the starting level to the second (flat) level.
To determine the mass rate of change of the space ship (Δm/Δt) needed to achieve an upward acceleration of 2.50 m/s², we can use the rocket equation, which states:
Δv = (ve * ln(m0 / mf))
Where:
Δv is the desired change in velocity (2.50 m/s² in the upward direction),
ve is the exhaust velocity of the fuel (40,000 m/s),
m0 is the initial mass of the space probe (190,000 kg + fuel mass),
mf is the final mass of the space probe (190,000 kg).
Rearranging the equation, we get:
Δm = m0 - mf = m0 * (1 - e^(Δv / ve))
To find the mass rate of change, we divide Δm by the time it takes to achieve the desired acceleration:
(Δm / Δt) = (m0 * (1 - e^(Δv / ve))) / t
To determine the vertical displacement of the roller coaster from its starting level when it reaches the second (flat) level with a velocity of 7.0 m/s, we can use the conservation of mechanical energy. At the starting level, the only form of energy is the potential energy stored in the compressed spring, which is then converted into kinetic energy at the second level.
Potential energy at the starting level = Kinetic energy at the second level
0.5 * k * x^2 = 0.5 * m * v^2
where:
k is the spring constant (550.0 N/m),
x is the compression of the spring (11.0 m),
m is the mass of the roller coaster cart (114.0 kg),
v is the velocity at the second level (7.0 m/s).
Plugging in the values:
0.5 * (550.0 N/m) * (11.0 m)^2 = 0.5 * (114.0 kg) * (7.0 m/s)^2
Solving this equation will give us the vertical displacement from the starting level to the second (flat) level.
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A horizontal force of 40lbs acts on a 12lb block moving it along a horizontal surface. The coefficient of static friction for the two surfaces in contact is 0.40, while the coefficient of kinetic friction for surfaces is 0.25. What is the acceleration of the block as it moves
along the surface?
The acceleration of the block as it moves along the surface is approximately 0.880 m/s².
To determine the acceleration of the block, we need to consider the forces acting on it.
The applied force:The horizontal force of 40 lbs (pounds) is acting on the block in the direction of motion.
The force of friction:The frictional force opposes the motion of the block. There are two cases we need to consider:
a) When the block is at rest (static friction):The maximum static friction force can be calculated using the formula:
F_static_max = μ_static * N
where μ_static is the coefficient of static friction and N is the normal force.
The normal force is equal to the weight of the block, which can be calculated as:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity.
b) When the block is in motion (kinetic friction):The kinetic friction force can be calculated using the formula:
F_kinetic = μ_kinetic * N
where μ_kinetic is the coefficient of kinetic friction and N is the normal force.
Once we have the forces, we can use Newton's second law to determine the acceleration:
ΣF = m * a
where ΣF is the net force acting on the block, m is the mass of the block, and a is the acceleration.
Applied force = 40 lbs
Mass of the block (m) = 12 lbs
Coefficient of static friction (μ_static) = 0.40
Coefficient of kinetic friction (μ_kinetic) = 0.25
Acceleration due to gravity (g) = 32.2 ft/s²
First, let's convert the values to SI units (kilograms and meters):
1 lb ≈ 0.454 kg
1 ft ≈ 0.305 m
Applying the conversions:Applied force = 40 lbs ≈ 40 * 0.454 kg ≈ 18.16 kg
Mass of the block (m) = 12 lbs ≈ 12 * 0.454 kg ≈ 5.448 kg
Acceleration due to gravity (g) = 32.2 ft/s² ≈ 32.2 * 0.305 m/s² ≈ 9.817 m/s²
Now, let's calculate the forces:
Static friction force:
N = m * g = 5.448 kg * 9.817 m/s² ≈ 53.467 N
F_static_max = μ_static * N = 0.40 * 53.467 N ≈ 21.387 N
Kinetic friction force:F_kinetic = μ_kinetic * N = 0.25 * 53.467 N ≈ 13.367 N
Since the applied force (40 lbs or 18.16 kg) exceeds the maximum static friction force (21.387 N), the block will start moving, and the kinetic friction force will be in effect. Therefore, the net force acting on the block is the difference between the applied force and the kinetic friction force:
ΣF = Applied force - F_kinetic = 18.16 kg - 13.367 N ≈ 4.793 N
Finally, we can use Newton's second law to calculate the acceleration:
ΣF = m * a
4.793 N = 5.448 kg * a
Solving for a:a ≈ 4.793 N / 5.448 kg ≈ 0.880 m/s²
Therefore, the acceleration of the block as it moves along the surface is approximately 0.880 m/s².
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1. An oil drop is balanced in a Millikan apparatus. The drop has a mass of 1.8 10-18 kg. The plates have a potential difference of 920 V, are separated by 3.6 cm, and the lower plate is positive. Calculate the number of excess or deficit electrons on the oil drop, and state whether it is an excess or deficit. [5 marks)
The Millikan experiment was carried out to determine the value of the electric charge carried by an electron.'
The method was to suspend oil droplets in a uniform electric field between two metal plates by adjusting the voltage applied to the plates such that the force on the droplet was balanced by the force of gravity. The excess or deficit charge on the droplet could then be calculated and from this,
The charge carried by an electron could be determined.What is an oil drop?An oil drop is a charged droplet of oil that is formed in a high voltage field. An oil droplet carries an electric charge because when it comes into contact with an ion.
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Answer the question with a cross in the box you think is correct. If you change your mind about an answer, put a line through the box and then mark your new answer with a cross When a guitar string is plucked, a sound of constant frequency is heard. The wave produced on the vibrating guitar string is A. longitudinal and progressive. B. longitudinal and stationary C. transverse and progressive. D. transverse and stationary
The wave produced on the vibrating guitar string is transverse and progressive.
When a guitar string is plucked, it produces a wave that travels along the string. This wave is transverse in nature, meaning that the particles of the medium (the string) vibrate perpendicular to the direction of wave propagation. As the string oscillates up and down, it creates peaks and troughs in the wave pattern, forming a characteristic waveform.
The wave is also progressive, which means it propagates through space. As the plucked string vibrates, the disturbance travels along the length of the string, carrying the energy of the wave with it. This progressive motion allows the sound wave to reach our ears, where we perceive it as a sound of constant frequency.
In summary, when a guitar string is plucked, it generates a transverse and progressive wave. The transverse nature of the wave arises from the perpendicular vibrations of the string's particles, while its progressiveness refers to the propagation of the wave through space, enabling us to hear a sound of constant frequency.
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Your answers are saved automatically. Remaining Time: 24 minutes, 55 seconds. Question completion Status: Moving to another question will save this response. Question 1 of 5 Question 1 0.5 points Save
The statement "[11] and [..] are linearly independent in M2.2" is false, the vectors are linearly dependent.
In order to determine if two vectors are linearly independent, we need to check if one vector can be expressed as a scalar multiple of the other vector. If it can, then otherwise, they are linearly independent.
Here, [11] and [..] are 2x2 matrices. The first vector [11] represents the matrix with elements 1 and 1 in the first row and first column, respectively. The second vector [..] represents a matrix with elements unknown or unspecified.
Since we don't have specific values for the elements in the second vector, we cannot determine if it can be expressed as a scalar multiple of the first vector. Without this information, we cannot definitively say whether the vectors are linearly independent or not. Therefore, the statement is false.
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The complete question is
Your answers are saved automatically Remaining Time: 24 minutes, 55 seconds. Question Completion Status: Moving to another question will save this response Question 1 of 5 Question 1 0.5 points Save of [11] [11] and [..] are linearly independent in M2.2 True False Moving to another question will save this response.
a) Show, starting with the First Law of Thermodynamics for a closed system, that, at constant pressure, the following form of the First Law is applicable to a closed system, AH=QH-W₂ where AH is the enthalpy change, QH is the heat energy supplied to the system and W, is the non-pV work done by the system. [20%] (b) A piston/cylinder system contains 10 kg of a wet steam mixture with a quality of 10.5% at a pressure of 5.00 bar. Using the following extract from steam Tables determine the specific enthalpy of the wet steam mixture in the system. [20%] (c) If 21.52 MJ is added as heat, and the pressure within the piston remains at 5.00 bar, what will be the final state of the system, i.e. what is the final temperature and what phase(s) will be present? [30%] (d) What are the volumes occupied by steam/water mixture described in (b), and the final volume of the system after the 21.52 MJ of heat has been added? [10%] (e) Determine the amount of pV work done by the system using (i) the volume change in the system and (ii) the internal energy change for the system. [20%]
The question involves deriving the First Law equation, determining specific enthalpy of a wet steam mixture, finding the final state of the system, calculating volumes and pV work, assessing thermodynamic principles and properties in a closed system.
What topics and calculations are involved in the given question related to the First Law of Thermodynamics, wet steam mixture, system states, and volume changes?The given question focuses on the application of the First Law of Thermodynamics for a closed system and involves various calculations related to enthalpy, heat energy, work, specific enthalpy, system states, and volume changes.
(a) In part (a), the derivation of the First Law of Thermodynamics at constant pressure is requested, showing the relationship AH = QH - W₂, where AH represents the enthalpy change, QH is the supplied heat energy, and W₂ is the non-pV work done by the system.
(b) In part (b), the specific enthalpy of a wet steam mixture is to be determined based on the provided information from steam tables.
(c) Part (c) involves determining the final state of the system, including the final temperature and the phases present, when a specific amount of heat is added while maintaining constant pressure.
(d) The volumes occupied by the initial steam/water mixture described in part (b) and the final volume of the system after the heat addition are requested in part (d).
(e) Part (e) requires the calculation of the pV work done by the system using two different approaches: the volume change in the system and the change in internal energy for the system.
Overall, the question assesses the understanding and application of thermodynamic principles and properties to analyze and solve problems related to energy, heat transfer, work, and system states in a closed system.
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The outside mirror on the piger side of a son and has focal length of sometive to the mirror a truck traveling in the rear has an object distance of time (a) Find the image distance of the truck m ASK Vind the magnification of the mirror
The outside mirror on the passenger side of a car is convex and has a focal length of- 7.0 m. Relative to this mirror, a truck traveling in the rear has an object distance of 11 m.(a)the image distance of the truck is approximately -4.28 meters.(b)the magnification of the convex mirror is approximately -0.389.
To find the image distance of the truck and the magnification of the convex mirror, we can use the mirror equation and the magnification formula.
Given:
Focal length of the convex mirror, f = -7.0 m (negative because it is a convex mirror)
Object distance, do = 11 m
a) Image distance of the truck (di):
The mirror equation is given by:
1/f = 1/do + 1/di
Substituting the given values into the equation:
1/(-7.0) = 1/11 + 1/di
Simplifying the equation:
-1/7.0 = (11 + di) / (11 × di)
Cross-multiplying:
-11 × di = 7.0 * (11 + di)
-11di = 77 + 7di
-11di - 7di = 77
-18di = 77
di = 77 / -18
di ≈ -4.28 m
The negative sign indicates that the image formed by the convex mirror is virtual.
Therefore, the image distance of the truck is approximately -4.28 meters.
b) Magnification of the mirror (m):
The magnification formula for mirrors is given by:
m = -di / do
Substituting the given values into the formula:
m = (-4.28 m) / (11 m)
Simplifying:
m ≈ -0.389
Therefore, the magnification of the convex mirror is approximately -0.389.
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A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.50 m/s. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction between the box and the surface is 0.20. i.) What horizontal force must the worker apply to maintain the constant motion? ii.) If the force is removed, how far does the box slide before coming to a rest? (HINT: In part, use kinematic expressions)
i) The worker must apply a horizontal force of 39.2 N to maintain the constant motion. ii) The box slides a distance of 8.75 m before coming to a rest.
i) To maintain a constant speed, the applied force must balance the frictional force acting on the box. The frictional force can be calculated using the formula F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.
Given that the coefficient of static friction is 0.40, we can find the normal force N using the equation N = mg, where m is the mass of the box and g is the acceleration due to gravity.
N = (11.2 kg)(9.8 m/s2) = 109.76 N
The frictional force is then F_friction = (0.40)(109.76 N) = 43.904 N.
ii) When the force is removed, the box experiences a deceleration due to the kinetic friction. The deceleration can be calculated using the formula a = F_friction / m, where F_friction is the kinetic frictional force and m is the mass of the box.
Using the kinematic equation [tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as, where v is the final velocity, u is the initial velocity (3.50 m/s), a is the acceleration, and s is the distance traveled, we can solve for s.
0 = (3.50 m/s)2 + 2(-1.964 m/s2) * s
Simplifying the equation, we find s = 8.75 m.
Therefore, the box slides a distance of approximately 8.75 m before coming to a rest.
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An acrobat hangs by his hands from the middle of a tightly stretched horizontal wire so that the angle between the wire and the horizontal is 8.57 If the acrobat's mass is 79.5 kg, what is the tension
The tension in the wire is approximately 785.06 Newtons.
To find the tension in the wire, we can analyze the forces acting on the acrobat.
The weight of the acrobat can be represented by the force mg, where m is the mass of the acrobat and g is the acceleration due to gravity.
In this scenario, there are two vertical forces acting on the acrobat: the tension in the wire and the weight of the acrobat. These forces must balance each other to maintain equilibrium.
The tension in the wire can be split into horizontal and vertical components. The vertical component of the tension will counteract the weight of the acrobat, while the horizontal component will be balanced by the horizontal force of the wire.
Using trigonometry, we can determine that the vertical component of the tension is T * cosθ, where T is the tension in the wire and θ is the angle between the wire and the horizontal.
Setting up the equation for vertical equilibrium, we have:
T * cosθ = mg
Solving for T, the tension in the wire, we get:
T = mg / cosθ
Substituting the given values, we have:
T = (79.5 kg) * (9.8 m/s^2) / cos(8.57°)
Calculating the tension using this formula will give us the answer.
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Two transverse waves y1 = 2 sin(2ttt - itx) and y2 = 2 sin(2nt - TeX + Tt/3) are moving in the same direction. Find the resultant amplitude of the interference between these two waves.
Two transverse waves y1 = 2 sin(2ttt - itx) and y2 = 2 sin(2nt - TeX + Tt/3) are moving in the same direction. The resultant amplitude of the interference between the two waves is √(8 + 8cos(nx)).
To find the resultant amplitude of the interference between the two waves, we need to add their individual amplitudes. The given waves are:
y1 = 2 sin(2ωt - k1x)
y2 = 2 sin(2ωt - k2x + φ)
where ω is the angular frequency, t is the time, k1 and k2 are the wave numbers, x is the position, and φ is the phase difference.
Comparing the equations, we can see that the angular frequency ω is the same for both waves (2ωt term). However, the wave numbers and phase differences are different.
k1 = ω, which implies k1 = 2t
k2 = ω, which implies k2 = n
Using the formula for the resultant amplitude of two interfering waves, we have:
Resultant amplitude = √(A1^2 + A2^2 + 2A1A2cos(φ))
In this case, A1 = 2 and A2 = 2 (both waves have the same amplitude).
To find the phase difference φ, we equate the phase terms in the given wave equations:
-itx = -k2x + φ
-itx = -nx + φ
Since the waves are moving in the same direction, we can assume that the phase difference φ is constant and does not depend on x. Therefore, we can rewrite the equation as:
φ = -itx + nx
Since we don't have specific values for t and n, we cannot determine the exact value of the phase difference φ.
However, if we assume that t = 0, then the equation becomes:
φ = 0 + nx = nx
In this case, the phase difference φ is directly proportional to x.
Now we can calculate the resultant amplitude:
Resultant amplitude = √(A1^2 + A2^2 + 2A1A2cos(φ))
= √(2^2 + 2^2 + 2(2)(2)cos(nx))
= √(4 + 4 + 8cos(nx))
= √(8 + 8cos(nx))
Therefore, the resultant amplitude of the interference between the two waves is √(8 + 8cos(nx)).
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33. A naturally occurring isotope of hydrogen called tritium (hydrogen-3) has a half-life of 12.3 years. If a sample of tritium is one-sixty-fourth of its original amount, how much time has elapsed si
The time elapsed since the original amount of tritium is one-sixty-fourth of its original amount can be determined by using the concept of half-life.
Tritium has a half-life of 12.3 years, which means that in every 12.3-year period, half of the tritium atoms decay.
To find the time elapsed, we can determine the number of half-lives that have occurred. Since the sample is one-sixty-fourth of its original amount, it has undergone 6 half-lives because 2^6 = 64.
Each half-life corresponds to a time period of 12.3 years, so the total time elapsed is 6 times the half-life, which is 6 * 12.3 = 73.8 years.
Therefore, the time elapsed since the original amount of tritium is one-sixty-fourth of its original amount is 73.8 years.
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An RL circuit is composed of a 12 V battery, a 6.0 H inductor and a 0.050 Ohm resistor.
The switch is closed at t = 0
An RL circuit is composed of a 12 V battery, a 6.0 H inductor and a 0.050 Ohm resistor.
The switch is closed at t = 0
These are the options:
The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is zero.
The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V.
The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V.
The time constant is 2.0 minutes an
The correct option is : The time constant is 2.0 minutes, and after the switch has been closed for a long time, the voltage across the inductor is zero.
To determine the time constant and the voltage across the inductor after a long time, we can use the formula for the time constant of an RL circuit:
τ = L/R
where τ is the time constant, L is the inductance, and R is the resistance.
In this case, the inductance (L) is given as 6.0 H and the resistance (R) is given as 0.050 Ω.
Using the formula, we can calculate the time constant:
τ = 6.0 H / 0.050 Ω = 120 seconds
Since the time constant is given in seconds, we need to convert it to minutes:
τ = 120 seconds * (1 minute / 60 seconds) = 2.0 minutes
So, the correct option is:
The time constant is 2.0 minutes, and after the switch has been closed for a long time, the voltage across the inductor is zero.
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An object 0.858 cm tall is placed 15.0 cm to the left of a concave spherical mirror having a radius of curvature of 20.6 cm. a. How far from the surface of the mirror is the image? Give the absolute v
The image is formed at a distance of 425.77 cm to the left of the mirror. The absolute value of the image distance will be 425.77 cm.
It is given that ,Height of object h1 = 0.858 cm, Distance of object from mirror u = -15.0 cm, Radius of curvature R = -20.6 cm
Since the mirror is concave in shape, its radius of curvature will be negative. By applying the mirror formula, we have the ability to determine the distance at which the image is positioned relative to the mirror.
That is, 1/f = 1/v + 1/u where,
the focal length of the mirror is denoted by f, and
v is the distance of the image from the mirror.
Rearranging the equation, we get,
1/v = 1/f - 1/u
1/f = 1/R
Therefore, substituting the values in the above equation, we get,
1/v = 1/R - 1/u = 1/-20.6 - 1/-15 = -0.0485v = -20.6/-0.0485v = 425.77 cm
As the image is formed on the same side of the object, the image distance v is negative. Thus, the image is formed at a distance of 425.77 cm to the left of the mirror. The absolute value of the image distance will be 425.77 cm.
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(a) Show that the function V(x, t) = A sin(kx – wt) cannot be a solution of the time- dependent Schrödinger equation with zero potential. (Hint: Note that k and w are real quantities.) = (b) Show that the function V (1,t) = Aciſka-wt) = A(cos(kr - wt) + i sin(kx – wt)) can tAeilkr - ) be a solution of the time-dependent Schrödinger equation with zero potential. What is the condition on w and k for it to be a solution? =
For the time-dependent Schrödinger equation with zero potential and given wave function
V (x, t) = A sin (k x – w t),
its form is as follows:
Hψ(x, t) = I ħ (∂ψ/∂t)
And,
ψ(x, t) = A sin (k x – w t) ∂ψ/∂
t = -A kw cos (k x-w t)
Hence,
Hψ(x, t) = iħ∂ψ/∂
t = -iħAk^2w cos (k x-w t)
Thus, the above wave function
V (x, t) = A sin (k x – w t)
cannot be a solution of the time- dependent Schrödinger equation with zero potential.
For the time-dependent Schrödinger equation with zero potential and given wave function
V (x, t) = cos (k r - w t) + i sin (k x – w t)),
its form is as follows:
Hψ(x, t) = i ħ (∂ψ/∂t) And,
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Problem 1: Water (density equal to 1000 kg/m3) flows through a system of pipes that goes up a step. The water pressure is 140 kPa at the bottom of the step (point 1), the cross-sectional area of the pipe at the top of the step (point 2) is half that at the bottom of the step and the speed of the water at the bottom of the step is 1.20 m/s. The pressure at the top of the step is 120 kPa. Find the value of the height h? (10 points) 0 V,
We need to found Find the value of the height h . To find the height we use the Bernoulli's equation .
The data of the problem as follows:
Water density, ρ = 1000 kg/m³
Water pressure at point 1, p1 = 140 kPa
Pressure at point 2, p2 = 120 kPa
Cross-sectional area of pipe at point 1, A1 = A2
Water speed at point 1, v1 = 1.20 m/s
Height difference between the two points, h = ? We are required to determine the value of height h.
Using Bernoulli's equation, we can write: `p1 + 1/2 ρ v1² + ρ g h1 = p2 + 1/2 ρ v2² + ρ g h2`
Here, as we need to find the value of h, we need to rearrange the equation as follows:
`h = (p1 - p2)/(ρ g) - (1/2 v2² - 1/2 v1²)/g`
To find the value of h, we need to calculate all the individual values. Let's start with the value of v2.The cross-sectional area of the pipe at point 2, A2, is half of the area at point 1, A1.A2 = (1/2) A
1We know that `v = Q/A` (where Q is the volume flow rate and A is the cross-sectional area of the pipe).As the volume of water entering a pipe must equal the volume of water exiting the pipe, we have:
Q = A1 v1 = A2 v2
Putting the values of A2 and v1 in the above equation, we get:
A1 v1 = (1/2) A1 v2v2 = 2 v1
Now, we can calculate the value of h using the above formula:
`h = (p1 - p2)/(ρ g) - (1/2 v2² - 1/2 v1²)/g`
Putting the values, we get:
`h = (140 - 120)/(1000 × 9.81) - ((1/2) (2 × 1.20)² - (1/2) 1.20²)/9.81`
Simplifying the above equation, we get:
h ≈ 1.222 m
Therefore, the answer is that the height difference between the two points is 1.222 m (approx).
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A
20-g cylinder of radius 5.0 cm starts to rotate from rest, reaching
200 rpm in half a minute. Find the net torque acting on the
cylinder.
The net torque acting on the cylinder is approximately 0.031 N·m.
To find the net torque acting on the cylinder, we can use the rotational motion equation:
Torque (τ) = Moment of inertia (I) × Angular acceleration (α).
Given that the cylinder starts from rest and reaches 200 rpm (revolutions per minute) in half a minute, we can calculate the angular acceleration. First, we convert the angular velocity from rpm to radians per second (rad/s):
ω = (200 rpm) × (2π rad/1 min) × (1 min/60 s) = 20π rad/s.
The angular acceleration (α) can be calculated by dividing the change in angular velocity (Δω) by the time taken (Δt):
α = Δω/Δt = (20π rad/s - 0 rad/s)/(30 s - 0 s) = (20π/30) rad/s².
Next, we need to calculate the moment of inertia (I) for the cylinder. The moment of inertia of a solid cylinder rotating about its central axis is given by:
I = (1/2)mr²,
where m is the mass of the cylinder and r is its radius.
Converting the mass of the cylinder from grams to kilograms, we have:
m = 20 g = 0.02 kg.
Substituting the values of m and r into the moment of inertia equation, we get:
I = (1/2)(0.02 kg)(0.05 m)² = 2.5 × 10⁻⁵ kg·m².
Now, we can calculate the net torque by multiplying the moment of inertia (I) by the angular acceleration (α):
τ = I × α = (2.5 × 10⁻⁵ kg·m²) × (20π/30) rad/s² ≈ 0.031 N·m.
Therefore, the net torque acting on the cylinder is approximately 0.031 N·m.
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