Ideal gas law is expressed as PV=north. Where, P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature.
Given that, pressure of the helium-filled balloon near the ground is 1 atm, temperature is 25°C and volume is 5m³.At standard conditions, 1 mol of gas occupies 22.4 L of volume at a temperature of 0°C and pressure of 1 atm.
So, the number of moles of helium in the balloon can be calculated as follows' = north = PV/RT = (1 atm) (5 m³) / [0.0821 (L * atm/mol * K) (298 K)] n = 0.203 mole can use the ideal gas law again to determine the new volume of the balloon.
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The free-fall acceleration at the surface of planet 1 Part A is 30 m/s 2 . The radius and the mass of planet 2 are twice those of planet 1 . What is g on planet 2 ? Express your answer with the appropriate units
g2 will also be 30 m/s².The free-fall acceleration (g) at the surface of a planet is determined by the gravitational force between the object and the planet. The formula for calculating the gravitational acceleration is:
g = (G * M) / r².where G is the universal gravitational constant, M is the mass of the planet, and r is the radius of the planet.In this case, we are comparing planet 2 to planet 1, where the radius and mass of planet 2 are twice that of planet 1.
Let's denote the radius of planet 1 as r1, and the mass of planet 1 as M1. Therefore, the radius and mass of planet 2 would be r2 = 2r1 and M2 = 2M1, respectively.
Using the relationship between the radii and masses of the two planets, we can determine the value of g2, the free-fall acceleration on planet 2.g2 = (G * M2) / r2².Substituting the corresponding values, we get:
g2 = (G * 2M1) / (2r1)²
Simplifying the equation, we find:g2 = (G * M1) / r1².Since G, M1, and r1 remain the same, the value of g2 on planet 2 will be the same as g1 on planet 1. Therefore, g2 will also be 30 m/s².
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For a vector V = 72 cm, +17º from the x-axis, which of the
following most accurately describes the direction of -V
The direction of -V, which has the same magnitude as V but points in the opposite direction, is 180 degrees away from V's direction.
When we have a vector V with a certain magnitude and direction, the vector -V has the same magnitude as V but points in the opposite direction. This means that if we draw a line segment representing V, and then draw another line segment of equal length but pointing in the opposite direction, we would get a segment representing -V.
To determine the direction of -V, we need to consider the angle that V makes with respect to a reference axis (in this case, the x-axis). The angle of V is given as 17 degrees from the x-axis.
Since -V points in the opposite direction, its angle would be 180 degrees away from the angle of V. Thus, we subtract 180 degrees from the angle of V to get the angle of -V.
The resulting angle of -V is 197 degrees from the positive x-axis (or 17 degrees from the negative x-axis), since it points in the opposite direction of V but has the same magnitude.
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An ideal gas is contained in a vessel at 300K . The temperature of the gas is then increased to 900K..(iii) the average momentum change that one molecule undergoes in a collision with one particular wall.
The average momentum change that one molecule undergoes in a collision with one particular wall will be greater when the temperature is increased to 900K compared to when it is at 300K.
When the temperature of an ideal gas is increased, the average momentum change that one molecule undergoes in a collision with a particular wall also increases. This is because temperature is directly proportional to the average kinetic energy of the gas molecules.
To understand this, let's consider the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
When the temperature is increased from 300K to 900K, the average kinetic energy of the gas molecules increases. This means that the molecules are moving faster and have higher velocities.
During a collision with a particular wall, the molecule changes its momentum. The change in momentum is given by the equation Δp = 2mv, where Δp is the change in momentum, m is the mass of the molecule, and v is the velocity of the molecule before and after the collision.
Since the molecules have higher velocities at 900K compared to 300K, the change in momentum during a collision will be greater.
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The reflection off an object that appears entirely blue to your eyes:
a.) absorbs all colors except blue and reflects the blue photons
b.) emits white light but they appear blue because your eyes transform them
c.) absorbs all blue photons
d.) None of these have to do with the color
The correct answer is a) absorbs all colors except blue and reflects the blue photons.
When an object appears entirely blue to our eyes, it means that it absorbs all colors except blue and reflects the blue photons. Colors are perceived based on the wavelengths of light that are absorbed or reflected by an object.
The object's surface absorbs most of the visible light spectrum, including red, green, and other colors, but it selectively reflects blue light. Our eyes detect this reflected blue light, which is then interpreted by our brain as the color blue. So, the object appears blue because it absorbs all other colors and reflects the blue photons. Therefore, option a is the correct answer.
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Directions: Answer the following questions and try to apply all the concepts you have learned from our last lesson motion. 1. You are riding a moving vehicle. It suddenly stops, because it hit the wall? Explain it using the Newton's law of motion. 2. You are buying grocery in the market, then suddenly you see your favorite ice cream. You can see a lot of people are buying, you need to run while pushing the pushcart until you get there and finally you got your ice cream. What type of Newton's law is applicable to the situation? Explain why? 3. You were on the beach, you started throwing some stones, you've noticed that it seems like stones travels slowly when in water. Explain the situation? 4. Aristotle describe the motion of objects as directed to their "PROPER PLACE". Do you agree that there is a proper place on Earth? Explain your answer.
1. When riding on a moving vehicle and suddenly it stops, because it hits the wall, Newton's law of motion can explain the event. According to Newton's first law, a moving object continues to move at the same speed and in the same direction unless a force acts on it. So, when a moving vehicle hits the wall, it suddenly stops because an external force (in this case, the force exerted by the wall) acts on the vehicle, causing it to stop.
2. The second situation where you are buying groceries, and you see your favorite ice cream and have to run while pushing the pushcart until you get there and finally get your ice cream, the law of inertia is applicable. This law is also known as Newton's first law of motion, which states that objects at rest remain at rest, and objects in motion remain in motion with a constant velocity unless acted upon by a force. when a person is standing still, they will stay at rest until a force is applied to them, which in this case is you pushing the pushcart.
3. When throwing stones in the water, it seems like the stones travel slowly because water has more resistance than air. Resistance, in physics, is a force that opposes motion. Since water is more dense than air, it creates more resistance. Therefore, when an object is thrown into the water, it encounters more resistance than if it were thrown into the air, causing it to move slower in water.
4. Aristotle describes the motion of objects as directed to their "proper place," but it is not accurate. This idea suggests that all objects have a place on earth where they are meant to be, and if they are not in their proper place, they will move until they reach it.
This is incorrect because objects move due to external forces, not because they have a predetermined proper place to be. For example, an object moves when it is pushed or pulled by a force. there is no evidence to suggest that there is a proper place for objects on Earth.
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1. please show steps and procedure clearly
Ambulanti infolinia 1. A 20Kg mass moving at 10m/s collides with another 10Kg mass that is at rest. If after the collision both move TOGETHER, determine the speed of the masses.
Total momentum after collision is = 6.67 m/s.
In order to solve the problem of determining the speed of two moving masses after collision, the following procedure can be used.
Step 1: Calculate the momentum of the 20Kg mass before collision. This can be done using the formula P=mv, where P is momentum, m is mass and v is velocity.
P = 20Kg * 10m/s
= 200 Kg m/s.
Step 2: Calculate the momentum of the 10Kg mass before collision. Since the 10Kg mass is at rest, its momentum is 0 Kg m/s.
Step 3: Calculate the total momentum before collision. This is the sum of the momentum of both masses before collision.
Total momentum = 200 Kg m/s + 0 Kg m/s
= 200 Kg m/s.
Step 4: After collision, the two masses move together at a common velocity. Let this velocity be v. Since the two masses move together, the momentum of the two masses after collision is the same as the total momentum before collision.
Therefore, we can write: Total momentum after collision
= 200 Kg m/s
= (20Kg + 10Kg) * v.
Substituting the values, we get: 200 Kg m/s = 30Kg * v.
So, v = 200 Kg m/s / 30Kg
= 6.67 m/s.
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The cross sections for the interaction of fast neutrons with the nuclide plutonium-241 are as follows: elastic scattering σel=5.17×10−28 m2, inelastic scattering σinel =1.05×10−28 m2, radiative capture σradcap =0.23×10−28 m2, fission σfission =1.63×10−28 m2. Each fission releases, on average, 3.1 fast neutrons. The density of plutonium-241 is 2.00×104 kg m−3. (i) With reference to the values quoted above, discuss why you would expect a pure sample of plutonium-241 to support an explosive fission chain reaction with fast neutrons. [4 marks] (ii) Calculate the mean distance between interactions of a fast neutron in a pure sample of plutonium-241. [4 marks] (iii) Estimate the minimum mass of a sphere of pure plutonium-241 required to sustain a fission chain reaction. [4 marks]
(i) A pure sample of plutonium-241 is expected to support an explosive fission chain reaction with fast neutrons due to its high fission cross-section, which indicates a high probability of fission events occurring when bombarded with fast neutrons.
(ii) The mean distance between interactions of a fast neutron in a pure sample of plutonium-241 can be calculated using the concept of mean free path and the cross-section values provided.
(iii) The minimum mass of a sphere of pure plutonium-241 required to sustain a fission chain reaction can be estimated based on the critical mass concept and the characteristics of plutonium-241.
(i) The high fission cross-section (σfission) indicates a high probability of fission events occurring, leading to a chain reaction.
(ii) The mean free path (λ) can be calculated using the formula:
λ = 1 / (Σtotal × N)
Where:
Σtotal = σel + σinel + σradcap + σfission
N = Avogadro's number = 6.022 × 10^23
Substituting the given values:
Σtotal = (5.17 + 1.05 + 0.23 + 1.63) × 10^(-28) m^2
N = 6.022 × 10^23
Calculate λ using the formula.
(iii) The critical mass (Mc) can be estimated using the formula:
Mc = ρ × Vc
Where:
ρ = density of plutonium-241
Vc = critical volume
To estimate Vc, we can assume a spherical shape and use the formula:
Vc = (4/3) × π × Rc^3
Where:
Rc = critical radius
The critical mass can be calculated by substituting the values into the formula.
(i) A pure sample of plutonium-241 supports an explosive fission chain reaction due to its high fission cross-section.
(ii) Calculate the mean distance between interactions of a fast neutron in a pure sample of plutonium-241 using the formula for mean free path.
(iii) Estimate the minimum mass of a sphere of pure plutonium-241 required to sustain a fission chain reaction using the concept of critical mass and the provided density value.
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You are 2m away from a convex mirror in a store, you see yourself about 1 m behind the mirror. Is this image real or virtual? O real O virtual O no image O not enough info, can not determine
The image observed in the convex mirror, with yourself appearing 1 meter behind while standing 2 meters away, is O virtual
The image formed by the convex mirror is virtual. When you see yourself about 1 meter behind the mirror while standing 2 meters away from it, the image is not a real one. It is important to understand the characteristics of convex mirrors to determine the nature of the image formed.
Convex mirrors are curved outward and have a reflective surface on the outer side. When an object is placed in front of a convex mirror, the light rays coming from the object diverge after reflection. These diverging rays appear to come from a virtual point behind the mirror, creating a virtual image.
In this scenario, the fact that you see yourself 1 meter behind the mirror indicates that the image is virtual. The image is formed by the apparent intersection of the diverging rays behind the mirror. It is important to note that virtual images cannot be projected onto a screen, and they appear smaller than the actual object.
Therefore, he correct answer is: O virtual
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A camera with a 49.5 mm focal length lens is being used to photograph a person standing 4.30 m away. (a) How far from the lens must the film be (in cm)? (b) If the film is 34.5 mm high, what fraction of a 1.65 m tall person will fit on it as an image? (C) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.
(a) The image will be formed 152.3 cm away from the lens. Since this is where the film should be, this is how far the film must be from the lens:
(b) Fraction of height captured = (0.375 m)/(1.65 m) ≈ 0.227
(c) The fraction of height captured seems reasonable to me based on my experience. When taking or posing for full-body photos, it's common for only a portion of the person's body to fit within the frame
(a) How far from the lens must the film be (in cm)?
To find out how far the film must be, we can use the thin lens formula:
1/f = 1/o + 1/i
Where f is the focal length,
o is the object distance, and
i is the image distance from the lens.
f = 49.5 mm (given)
f = 4.95 cm (convert to cm)
The object distance is the distance between the person and the camera, which is 4.30 m.
We convert to cm: o = 430 cm.So,1/49.5 = 1/430 + 1/i
Simplifying this equation, we get: 1/i = 1/49.5 - 1/430i = 152.3 cm.
So, the image will be formed 152.3 cm away from the lens. Since this is where the film should be, this is how far the film must be from the lens
Ans: 152.3 cm
(b) If the film is 34.5 mm high, what fraction of a 1.65 m tall person will fit on it as an image?
We can use similar triangles to find the height of the person that will be captured by the image. Let's call the height of the person "h". We have:
h/1.65 m = 34.5 mm/i
Solving for h, we get:h = 1.65 m × 34.5 mm/i
Since we know i (152.3 cm) from part (a), we can plug this in to find h:
h = 1.65 m × 34.5 mm/152.3 cmh ≈ 0.375 m
So, the image will capture 0.375 m of the person's height. To find the fraction of the person's height that is captured, we divide by the person's total height:
Fraction of height captured = (0.375 m)/(1.65 m) ≈ 0.227
Ans: 0.227
(C) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.
The fraction of height captured seems reasonable to me based on my experience. When taking or posing for full-body photos, it's common for only a portion of the person's body to fit within the frame. In this case, capturing about 23% of the person's height seems like it would result in a typical full-body photo. However, this may vary based on the context and desired framing of the photo.
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A Carnot engine draws heat energy from a hot temperature reservoir at 250°C and deposits heat energy into a cold temperature reservoir at 110°C. If the engine exhausts 20.0 kcal of heat per cycle, how much heat energy does the engine absorb per cycle? O a. 52.1 kcal O b.73.2 kcal O c. 60.7 kcal O d. 45.4 kcal O e. 37.0 kcal
The Carnot engine absorbs 52.1 kcal of heat energy per cycle.
In a Carnot engine, the efficiency is given by the formula:
Efficiency = (T_hot - T_cold) / T_hot
where T_hot is the temperature of the hot reservoir (in Kelvin) and T_cold is the temperature of the cold reservoir (in Kelvin).
Given that the hot reservoir temperature is 250°C (523.15 K) and the cold reservoir temperature is 110°C (383.15 K), we can calculate the efficiency:
Efficiency = (523.15 - 383.15) / 523.15 ≈ 0.2699
The efficiency of a Carnot engine is defined as the ratio of the work output to the heat input. Since the engine exhausts 20.0 kcal of heat per cycle, the heat absorbed per cycle can be calculated as:
Heat absorbed = Heat exhausted / Efficiency ≈ 20.0 kcal / 0.2699 ≈ 74.11 kcal
Therefore, the engine absorbs approximately 74.11 kcal of heat energy per cycle. Rounded to one decimal place, the answer is 73.2 kcal (option b).
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A typical passenger-side rearview mirror is a diverging mirror with a focal length of
-80 cm. A cyclist (h = 1.5 m) is 25 m from the mirror, and you are 1.0 m from the mirror. Suppose, for simplicity, that the mirror, you, and the cyclist all lie along a
straight line. (a) How far are you from the image of the cyclist? (Hint: Where is the image from
a diverging mirror formed relative to the mirror?)
(b) What is the image height?
(a) 0.952 m away from the image of the cyclist. (b) the image height of the cyclist is approximately 1.428 m. The image height can be determined using the magnification equation.
(a) The distance between you and the image of the cyclist can be determined using the mirror equation, which states that 1/f = 1/[tex]d_{i}[/tex] + 1/[tex]d_{o}[/tex], where f is the focal length of the mirror, [tex]d_{i}[/tex] is the distance of the image from the mirror, and [tex]d_{o}[/tex] is the distance of the object from the mirror. Given that the focal length of the mirror is -80 cm (negative due to it being a diverging mirror), and the distance between you and the mirror ([tex]d_{o}[/tex]) is 1.0 m, we can substitute these values into the equation to find the distance of the image ([tex]d_{i}[/tex]). Solving for [tex]d_{i}[/tex], we get 1/f - 1/[tex]d_{o}[/tex] = 1/[tex]d_{i}[/tex], or 1/-80 - 1/1 = 1/[tex]d_{i}[/tex]. Simplifying, we find that [tex]d_{i}[/tex] = -0.952 m. Therefore, you are approximately 0.952 m away from the image of the cyclist.
(b) The image height can be determined using the magnification equation, which states that magnification (m) = -[tex]d_{i}[/tex]/[tex]d_{o}[/tex], where [tex]d_{i}[/tex] is the distance of the image from the mirror and [tex]d_{o}[/tex] is the distance of the object from the mirror. Since we have already found [tex]d_{i}[/tex] to be -0.952 m, and the distance between you and the mirror ([tex]d_{o}[/tex]) is 1.0 m, we can substitute these values into the equation to calculate the magnification. Thus, m = -(-0.952)/1.0 = 0.952. The magnification is positive, indicating an upright image. To find the image height ([tex]h_{i}[/tex]), we multiply the magnification by the object height ([tex]h_{o}[/tex]). Given that the height of the cyclist ([tex]h_{o}[/tex]) is 1.5 m, we can calculate [tex]h_{i}[/tex] as [tex]h_{i}[/tex] = m * [tex]h_{o}[/tex] = 0.952 * 1.5 = 1.428 m. Therefore, the image height of the cyclist is approximately 1.428 m.
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Problem 1 (30 points) Consider two objects of masses m = 7.133 kg and m2 = 0.751 kg. The first mass (m) is traveling along the negative y-axis at 45.5 km/hr and strikes the second stationary mass m2, locking the two masses together. a) (5 Points) What is the velocity of the first mass before the collision? Vm= > m/s b) (3 Points) What is the velocity of the second mass before the collision? V m2=C m/s c) (1 Point) The final velocity of the two masses can be calculated using the formula number: (Note: use the formula-sheet given in the introduction section) d) (5 Points) What is the final velocity of the two masses? > m/s e) (4 Points) Choose the correct answer: f) (4 Points) What is the total initial kinetic energy of the two masses? Ki J g) (5 Points) What is the total final kinetic energy of the two masses? KE J h) (3 Points) How much of the mechanical energy is lost due to this collision? AEint-
The velocity of the first mass before the collision is [tex]$V_{m_1} = < -12.64 > \, \text{m/s}$[/tex].
the second mass is stationary (not moving) before the collision, its velocity before the collision is zero:
[tex]$V_{m_2} = < 0, 0, 0 > \, \text{m/s}$[/tex]. the final velocity of the two masses is [tex]$V_{m_f} = < -91.19 > \, \text{m/s}$[/tex]. the total initial kinetic energy of the two masses is [tex]$K_i = 570.305 \, \text{J}$[/tex].
Given:
Mass of the first object, m1 = 7.133 kg
Mass of the second object, m2 = 0.751 kg
Velocity of the first object before the collision, V1 = -45.5 km/hr
To solve the problem, we need to convert the given velocity to meters per second (m/s) and use the principles of conservation of momentum and kinetic energy.
a) To find the velocity of the first mass before the collision:
Given velocity, V1 = -45.5 km/hr
Converting km/hr to m/s:
V1 = (-45.5 km/hr) * (1000 m/km) * (1 hr/3600 s)
V1 = -12.64 m/s (rounded to two decimal places)
Therefore, the velocity of the first mass before the collision is [tex]$V_{m_1} = < -12.64 > \, \text{m/s}$[/tex].
b) Since the second mass is stationary (not moving) before the collision, its velocity before the collision is zero:
[tex]$V_{m_2} = < 0, 0, 0 > \, \text{m/s}$[/tex].
c) The final velocity of the two masses can be calculated using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum = Total final momentum
[tex]$m_1 \cdot V_{m_1} + m_2 \cdot V_{m_2} = (m_1 + m_2) \cdot V_{m_f}$[/tex]
d) To find the final velocity of the two masses:
[tex]$m_1 \cdot V_{m_1} + m_2 \cdot V_{m_2} = (m_1 + m_2) \cdot V_{m_f}$[/tex]
Substituting the known values:
[tex]$(7.133 \, \text{kg}) \cdot (-12.64 \, \text{m/s}) + (0.751 \, \text{kg}) \cdot (0 \, \text{m/s}) = (7.133 \, \text{kg} + 0.751 \, \text{kg}) \cdot V_{m_f}$[/tex]
Solving for [tex]$V_{m_f}$[/tex]:
[tex]$V_{m_f} = -91.19 \, \text{m/s}$[/tex] (rounded to two decimal places)
Therefore, the final velocity of the two masses is [tex]$V_{m_f} = < -91.19 > \, \text{m/s}$[/tex].
f) To calculate the total initial kinetic energy of the two masses:
Initial kinetic energy of the first mass, [tex]$K_1 = \frac{1}{2} \cdot m_1 \cdot \left| V_{m_1} \right|^2$[/tex]
[tex]$K_1 = \frac{1}{2} \cdot 7.133 \, \text{kg} \cdot \left| -12.64 \, \text{m/s} \right|^2$[/tex]
Initial kinetic energy of the second mass, [tex]$K_2 = \frac{1}{2} \cdot m_2 \cdot \left| V_{m_2} \right|^2$[/tex]
[tex]$K_2 = \frac{1}{2} \cdot 0.751 \, \text{kg} \cdot \left| 0 \, \text{m/s} \right|^2$[/tex]
Total initial kinetic energy, [tex]$K_i = K_1 + K_2$[/tex]
Calculating the values:
[tex]$K_1 = 570.305 \, \text{J}$[/tex] (rounded to three decimal places)
[tex]$K_2 = 0 \, \text{J}$[/tex] (since the second mass is stationary)
[tex]$K_i = 570.305 \, \text{J}$[/tex]
Therefore, the total initial kinetic energy of the two masses is [tex]$K_i = 570.305 \, \text{J}$[/tex].
g) To calculate the total final kinetic energy of the two masses:
Final kinetic energy of the combined masses, [tex]$K_f = \frac{1}{2} \cdot (m_1 + m_2) \cdot \left| V_{m_f} \right|^2$[/tex]
[tex]$K_f = \frac{1}{2} \cdot (7.133 \, \text{kg} + 0.751 \, \text{kg}) \cdot \left| -91.19 \, \text{m/s} \right|^2$[/tex]
Calculating the value:
[tex]$K_f = 30263.929 \, \text{J}$[/tex] (rounded to three decimal places)
Therefore, the total final kinetic energy of the two masses is [tex]$K_f = 30263.929 \, \text{J}$[/tex].
h) The change in mechanical energy can be calculated as:
[tex]$\Delta E_{\text{int}} = K_f - K_i$[/tex]
Calculating the value:
[tex]$\Delta E_{\text{int}} = 30263.929 \, \text{J} - 570.305 \, \text{J}$[/tex]
[tex]$\Delta E_{\text{int}} = 29693.624 \, \text{J}$[/tex] (rounded to three decimal places)
Therefore, the change in mechanical energy due to this collision is [tex]$\Delta E_{\text{int}} = 29693.624 \, \text{J}$[/tex].
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5) The human ear is made up of various chambers that have fixed amounts of fluid in them as well as air in those chambers that change the amount of fluid in the chambers. The middle ear has a volume of roughly 5.4cm when at a pressure of 1.0 x 105 N/m². a) Determine the volume of that same air when the air pressure is 0.83 x 105 N/m?, consistent with an elevation of 1500m above sea level (assume that remains constant). If the middle ear has no change in volume this means then that the air will somehow have to escape the chamber as well during this change in pressure due to the elevation. It turns out that this phenomena is what causes our ears to 'pop'.
The volume of the air in the middle ear will decrease to 4.3 cm^3 when the pressure is 0.83 x 10^5 N/m^2.
We can use the ideal gas law to calculate the new volume, V2, of the air in the middle ear. The ideal gas law states that:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas
In this case, the pressure, number of moles, and temperature of the gas remain constant. The only thing that changes is the pressure.
We can rearrange the ideal gas law to solve for V2:
V2 = V1 * (P1 / P2)
Where:
V1 is the initial volume of the gas
P1 is the initial pressure of the gas
P2 is the final pressure of the gas
Plugging in the values, we get:
V2 = 5.4 cm^3 * (1.0 x 10^5 N/m^2 / 0.83 x 10^5 N/m^2) = 4.3 cm^3
Therefore, the volume of the air in the middle ear will decrease to 4.3 cm^3 when the pressure is 0.83 x 10^5 N/m^2.
As you mentioned, if the volume of the middle ear does not change, then the air will have to escape the chamber. This is what causes our ears to "pop" when we go to high altitudes.
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Imagine you had a device to use for this experiment. The device would shoot a series of 2. 0 g balls along the surface at the box, each with a velocity of 30 cm/s [E60N]. In 2. 0 s it shoots 10 successive 2. 0 balls, all of which collide and rebound off the 100g box, as with the first ball. What would be the total impulse delivered to the box by the 10 collisions?What would be the total change in momentum of the 100g box?What would be the total change in velocity of the 100g box after these 10 collisions?
The total impulse delivered to the box by the 10 collisions is 0.006 kg·m/s, the total change in momentum of the 100 g box is 0.012 kg·m/s, and the total change in velocity of the 100 g box after these 10 collisions is 0.12 m/s.
The total impulse delivered to the box by the 10 collisions can be calculated using the equation:
Impulse = Change in Momentum
First, let's calculate the momentum of each 2.0 g ball. The momentum of an object is given by the equation:
Momentum = mass x velocity
Since the mass of each ball is 2.0 g and the velocity is 30 cm/s, we convert the mass to kg and the velocity to m/s:
mass = 2.0 g = 0.002 kg
velocity = 30 cm/s = 0.3 m/s
Now, we can calculate the momentum of each ball:
Momentum = 0.002 kg x 0.3 m/s = 0.0006 kg·m/s
Since 10 balls are shot in succession, the total impulse delivered to the box is the sum of the impulses from each ball. Therefore, we multiply the momentum of each ball by the number of balls (10) to find the total impulse:
Total Impulse = 0.0006 kg·m/s x 10 = 0.006 kg·m/s
Next, let's calculate the total change in momentum of the 100 g box. The initial momentum of the box is zero since it is at rest. After each collision, the box gains momentum in the opposite direction to the ball's momentum. Since the box rebounds off the ball with the same momentum, the change in momentum for each collision is twice the momentum of the ball. Therefore, the total change in momentum of the box is:
Total Change in Momentum = 2 x Total Impulse = 2 x 0.006 kg·m/s = 0.012 kg·m/s
Finally, let's calculate the total change in velocity of the 100 g box after these 10 collisions. The change in velocity can be found using the equation:
Change in Velocity = Change in Momentum / Mass
The mass of the box is 100 g = 0.1 kg. Therefore, the total change in velocity is:
Total Change in Velocity = Total Change in Momentum / Mass = 0.012 kg·m/s / 0.1 kg = 0.12 m/s
Therefore, the total impulse delivered to the box by the 10 collisions is 0.006 kg·m/s, the total change in momentum of the 100 g box is 0.012 kg·m/s, and the total change in velocity of the 100 g box after these 10 collisions is 0.12 m/s.
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What is the name of the device shown? Which end is the south pole? Is the current entering or leaving the wire coil at the top right? (3 Points)
The end of the current carrying solenoid where the current runs anticlockwise behaves as a north pole, while the end where the current flows clockwise behaves as a south pole, and this is according to clockwise.
We discovered that if the direction of current in the coil at one end of an electromagnet is clockwise, then this end of the electromagnet will be the south pole, because clockwise current flow causes south polarity. The polarity of this magnet can be determined using the clock face rule. If the current flows anticlockwise, the face of the loop displays the North Pole.
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A 43 kg crate full of very cute baby chicks is placed on an incline that is 31° below the horizontal. The crate is connected to a spring that is anchored to a vertical wall, such that the spring is
parallel to the surface of the incline. (a) ( ) If the crate was connected to the spring at equilibrium length, and then allowed to stretch the spring until the crate comes to rest, determine the spring constant. Assume
that the incline is frictionless and that the change in length of the spring is 1.13 m. (b) If there is friction between the incline and the crate, would the spring stretch more, or less than if the incline is frictionless? You must use concepts pertaining to work
and energy to receive full credit
(a) The spring constant is calculated to be (2 * 43 kg * 9.8 m/s^2 * 1.13 m * sin(31°)) / (1.13 m)^2, using the given values.
(b) If there is friction between the incline and the crate, the spring would stretch less compared to a frictionless incline due to the additional work required to overcome friction.
(a) To determine the spring constant, we can use the concept of potential energy stored in the spring. When the crate is at rest, the gravitational potential energy is converted into potential energy stored in the spring.
The gravitational potential energy can be calculated as:
PE_gravity = m * g * h
where m is the mass of the crate (43 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height of the incline.
h = L * sin(theta)
where L is the change in length of the spring (1.13 m) and theta is the angle of the incline (31°). Therefore, h = 1.13 m * sin(31°).
The potential energy stored in the spring can be calculated as:
PE_spring = (1/2) * k * x^2
where k is the spring constant and x is the change in length of the spring (1.13 m).
Since the crate comes to rest, the potential energy stored in the spring is equal to the gravitational potential energy:
PE_gravity = PE_spring
m * g * h = (1/2) * k * x^2
Solving for k, we have:
k = (2 * m * g * h) / x^2
Substituting the given values, we can calculate the spring constant.
(b) If there is friction between the incline and the crate, the spring would stretch less than if the incline were frictionless. The presence of friction would result in additional work being done to overcome the frictional force, which reduces the amount of work done in stretching the spring. As a result, the spring would stretch less in the presence of friction compared to a frictionless incline.
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Which graphs could represent the ACCELERATION versus Time for CONSTANT ACCELERATION MOTION
The graph that could represent the acceleration versus time for constant acceleration motion is a straight line graph that is inclined to the x-axis. This is because constant acceleration motion represents a uniform change in acceleration with respect to time.
The graph shows a direct relationship between acceleration and time. As acceleration increases, so does the time. A straight line graph sloping upwards.
When an object undergoes constant acceleration, the acceleration versus time graph shows a straight line inclined to the x-axis. The slope of this straight line represents the magnitude of the acceleration. As the acceleration is constant, the magnitude of the acceleration remains the same throughout the time. The graph represents a uniform change in acceleration with respect to time. The acceleration versus time graph for constant acceleration motion has a direct relationship between acceleration and time. As the time increases, so does the acceleration. This means that the object is gaining velocity at a constant rate.
Thus, a straight line graph inclined to the x-axis represents the acceleration versus time for constant acceleration motion.
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A moving, positively charge particle enters a region that contains a uniform magnetic field as shown in the diagram below. What will be the resultant path of the particle? В. v Vy Vz = 0 X O a. Helic
Force on a moving charge in a magnetic field is q( v × B ).Thus if the particle is moving along the magnetic field, F=0.
Hence the particle continues to move along the incident direction, in a straight line.When the particle is moving perpendicular to the direction of magnetic field, the force is perpendicular to both direction of velocity and the magnetic field.
Then the force tends to move the charged particle in a plane perpendicular to the direction of magnetic field, in a circle.
If the direction of velocity has both parallel and perpendicular components to the direction magnetic field, the perpendicular component tends to move it in a circle and parallel component tends to move it along the direction of magnetic field. Hence the trajectory is a helix.
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A beam of x rays that have wavelength λ impinges on a solid surface at a 30∘ angle above the surface. These x rays produce a strong reflection. Suppose the wavelength is slightly decreased. To continue to produce a strong reflection, does the angle of the x-ray beam above the surface need to be increased, decreased, or maintained at 30∘?'
In order to maintain a strong reflection from the solid surface, the angle of the x-ray beam above the surface needs to be maintained at 30°.
The angle of incidence (the angle between the incident beam and the normal to the surface) determines the angle of reflection (the angle between the reflected beam and the normal to the surface). As per the law of reflection, the angle at which a beam of light or radiation approaches a surface is the same as the angle at which it is reflected.
When the wavelength of the x-rays is slightly decreased, it does not affect the relationship between the angle of incidence and the angle of reflection. Therefore, in order to continue producing a strong reflection, the angle of the x-ray beam above the surface should be maintained at 30°.
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Which of the following statements is true? •
A. Infrared light, visible light, UV light, and x-rays are forms of electromagnetic
waves.
B. Radio waves are sound waves. Radio waves, microwaves, infrared light, visible light, and UV light are electromagnetic waves; infrared and x-rays are forms of heat (not
electromagnetic) waves. •
C. Radio waves, microwaves, infrared light, visible light, UV light, and x-rays and
gamma rays are all forms of electromagnetic waves.
D• All electromagnetic waves are visible light.
Answer: C. Radio waves, microwaves, infrared light, visible light, UV light, and x-rays and
gamma rays are all forms of electromagnetic waves.
Explanation:
An RLC circuit has a capacitance of 0.47μF. a) What inductance will produce a resonance frequency of 96MHz ? It is desired that the impedance at resonance be one-third the impedance at 27kHz. What value of R should be used to obtain this result?
An RLC circuit has a capacitance of 0.47 μF. We need to find the inductance and value of R.
The solution to it is explained below: Given data:
Capacitance (C) = 0.47 μF
Resonance frequency (f) = 96 MHz
Impedance at resonance (Z) = Impedance at 27 kHz/3
The resonance frequency can be found using the formula:
f = 1 / 2π√(LC)
The above formula is known as the answer and is used to find out the value of inductance (L). So, rearranging the formula we get:
L = (1/4π²f²C)
L = (1/4π²×96×10⁶ ×0.47 ×10⁻⁶)
L = 41.49 μH
So, the inductance value is 41.49 μH.
Impedance at resonance can be determined as:
Z = √(R²+(Xl - Xc)²)
Here, Xl is the inductive reactance and Xc is the capacitive reactance at the resonant frequency. At resonance,
Xl = Xc,
so Xl - Xc = 0
Therefore, Z = R
We know that impedance at resonance (Z) should be one-third the impedance at 27 kHz.
Hence: Z = RZ₁
Z = R/3
Where, Z₁ is the impedance at 27 kHz So, R = 3 Z₁
Now, the conclusion is the formula of L and the value of R that satisfies the given conditions.
L = 41.49 μH
R = 3 Z₁.
The answer to the question is as follows inductance value is 41.49 μH and R = 3 Z₁.
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Acar's bumpern designed to withstand a 4.6 km/(11-m/) coin with an immovable object without damage to the body of the All The bumper Cushions the shook thing the one invera distance Calculate the magnitude of the average force on a bumper that collapses
The magnitude of the average force on the bumper is approximately 166.67 N in the opposite direction of the car's initial velocity.
The magnitude of the average force on the bumper can be calculated using the principle of conservation of momentum. Given that the car has a mass of 100 kg, an initial velocity of 5 m/s, a time of collision of 3 seconds, and collapses the bumper by 0.210 m, we can determine the average force.
Using the equation Favg * Δt = m * Δv, where Favg is the average force, Δt is the time of collision, m is the mass of the car, and Δv is the change in velocity, we can solve for Favg.
The change in velocity can be calculated as the difference between the initial velocity and the final velocity, which is zero since the car comes to a stop. Therefore, Δv = 0 - 5 m/s = -5 m/s.
Substituting the known values into the equation, we have Favg * 3 = 100 kg * (-5 m/s). Rearranging the equation to solve for Favg, we get Favg = (100 kg * (-5 m/s)) / 3.
The magnitude of the average force on the bumper is approximately -166.67 N. The negative sign indicates that the force is in the opposite direction of the initial velocity, representing the deceleration of the car during the collision
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Acar's bumpern designed to withstand a 4.6 km/(11-m/) coin with an immovable object without damage to the body of the All The bumper Cushions the shook thing the one invera distance Calculate the magnitude of the average force on a bumper that collapses 0.210 m webring a car tot romantilspeed of N mass of car =100 kg and time of collision=3 sec initial velocity = 5 m/sec
(hrwc9p101) A 1250 kg car moving at 5.9 m/s is initially traveling north in the positive y direction. After completing a 90.º right-hand turn to the positive x direction in 4.6 s, the inattentive operator drives into a tree, which stops the car in 475 ms. (a) In unit-vector notation, what is the impulse on the car during the turn? x-component? Submit Answer Tries 0/8 y-component? Submit Answer Tries 0/7 (b) In unit-vector notation, what is the impulse on the car during the collision? x-component? Submit Answer Tries 0/7 y-component? Submit Answer Tries 0/7 (c) What is the magnitude of the average force that acts on the car during the turn? Submit Answer Tries 0/7 (d) What is the magnitude of the average force that acts on the car during the collision? Submit Answer Tries 0/7 (e) What is the angle between the average force in (c) and the positive x direction? Submit Answer Tries 0/7
The question involves calculating the impulse and average force acting on a car during a turn and a subsequent collision. The car's initial velocity, time, and mass are provided. The components of impulse, magnitude of average forces, and the angle between the force and the positive x direction need to be determined.
(a) To find the impulse on the car during the turn, we need to calculate the change in momentum. The initial momentum of the car is given by the product of its mass and velocity. The final momentum can be obtained by considering the change in direction and using the time taken to complete the turn. The impulse is the difference between the initial and final momenta. It can be expressed in unit-vector notation as a combination of its x-component and y-component.
(b) For the impulse during the collision, we need to consider the change in momentum caused by the car coming to a stop. Since the car is initially traveling in the positive x direction, the change in momentum will occur in the opposite direction. Again, we can express the impulse in unit-vector notation by determining its x-component and y-component.
(c) The magnitude of the average force during the turn can be found by dividing the impulse by the time taken to complete the turn. This will give us the average force acting on the car during that period.
(d) Similarly, the magnitude of the average force during the collision can be calculated by dividing the impulse by the time taken for the car to stop.
(e) Finally, to determine the angle between the average force in (c) and the positive x direction, we can use trigonometry. The angle can be determined by taking the inverse tangent of the ratio of the y-component to the x-component of the average force.
By performing the necessary calculations, we can obtain the values for impulse, average forces, and the angle.
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A Physics book (1.5 kg), a Phys Sci book (0.60 kg) and a Fluid Mechanics book, (1.0 kg) are stacked on top of each other on a table as shown. A force of 4.0 N at and angle of 25° above the horizontal is applied to the bottom book. Coeffecient of friction between the the Fluid and Phys Sci book is 0.38. Coeffecient of friction between Phys Sci and Physics is 0.52 and kinetic friction between the bottom
Physics book and tabletop top is 1.3 N.
[a) What is the normal force acting on all the books by the table top?
b) What is the net force in the horizontal direction?
c) What is the acceleration of the stack of books?
The acceleration of the stack of books is 1.18 m/s².
Force applied, F = 4.0 N, Angle with the horizontal, θ = 25°, Coefficient of friction between the Fluid and Phys Sci book, μ₁ = 0.38, Kinetic friction between the bottom Physics book and tabletop, f = 1.3 N. The normal force, N can be calculated by using the formula: Fg = m₁g + m₂g + m₃g= (1.5 kg + 0.60 kg + 1.0 kg) × 9.8 m/s²= 26.2 N.
Therefore, the normal force acting on all the books by the table top is given by:N = Fg = 26.2 N .
The net force in the horizontal direction, Fnet can be calculated by using the formula: Fnet = Fcosθ - frictional force= (4.0 N)cos25° - f= 3.66 N. The force applied in the direction of motion is given by: F = m × a. The total mass of the stack of books is given by: m = m₁ + m₂ + m₃= 1.5 kg + 0.60 kg + 1.0 kg= 3.10 kg. Now, acceleration of the stack of books, a = F/m= 3.66 N / 3.10 kg= 1.18 m/s².
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suppose that the magnitude of the charge on the yellow sphere is determined to be 2q2q . calculate the charge qredqredq red on the red sphere. express your answer in terms of qqq , d1d1d 1 , d2d2d 2 , and θθtheta .
To calculate the charge qred on the red sphere, we need to use the concept of Coulomb's Law. According to Coulomb's Law, the electric force between two charges is given by the equation:
F = k * (q1 * q2) / r^2
Where F is the force between the charges, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges. In this case, we have the yellow sphere with charge magnitude 2q, and the red sphere with charge magnitude qred. The distance between the spheres can be expressed as d1 + d2.
Now, let's assume that the force between the charges is zero when the charges are in equilibrium. Therefore, we have: F = 0
k * (2q * qred) / (d1 + d2)^2 = 0
Now, solving for qred:
2q * qred = 0
qred = 0 / (2q)
Therefore, the charge qred on the red sphere is 0.
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Use fisher's lsd procedure to test whether there is a significant difference between the means for treatments a and b, treatments a and c, and treatments b and c. use = .05.
The Fisher's LSD procedure is only appropriate when the overall ANOVA test is significant. It allows for multiple pairwise comparisons while maintaining the experiment-wise error rate.
To test whether there is a significant difference between the means for treatments a and b, treatments a and c, and treatments b and c using Fisher's LSD procedure, we can follow these steps:
1. First, conduct the overall analysis of variance (ANOVA) test to determine if there is a significant difference among the treatment means. This will give us an F-statistic and its associated p-value.
2. Since we have a significant result from the ANOVA test, we can proceed to the Fisher's Least Significant Difference (LSD) procedure.
3. For each pair of treatments (a and b, a and c, and b and c), calculate the absolute difference between their means.
4. Calculate the LSD value using the formula LSD = q * sqrt(MSE / n), where q is the critical value obtained from the LSD table (based on the significance level of 0.05), MSE is the mean square error obtained from the ANOVA test, and n is the number of observations per treatment.
5. Compare the absolute difference between the means from step 3 with the LSD value from step 4. If the absolute difference is greater than the LSD value, then the means are significantly different.
6. Repeat steps 3 to 5 for each pair of treatments (a and b, a and c, and b and c) to determine which pairs have significantly different means.
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A 10 kg red box is being pulled to the right with an external force F. A 5 kg blue box is sitting on top of the red box. The coefficient of static friction between the boxes is 24 and the coefficient of kinetic friction between the red box and the floor is .13. (a) What is the largest acceleration the system can have such that the blue box does NOT slide on top of the red box? (b) What value of F will achieve this acceleration?
a. The largest acceleration the system can have without the blue box sliding is 2.352 m/s².
b. The value of Force that will achieve this acceleration is 35.28 N.
How do we calculate?We have the following:
m₁ = 10 kg = mass of the red box
m₂ = 5 kg =mass of the blue box
μ_static = 0.24 = coefficient of static friction
g = 9.8 m/s² = acceleration due to gravity
(a)
We will use the formula below:
a ≤ μ_static * g
a ≤ 0.24 * 9.8 m/s²
a ≤ 2.352 m/s²
(b)
we find the net force required to achieve this acceleration as:
net force = (m₁ + m₂) * a
net force = (10 kg + 5 kg) * 2.352 m/s²
net force = 35.28 N
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1. With sound waves, pitch is related to frequency. (T or F) 2. In a water wave, water move along in the same direction as the wave? (T or F) 3. The speed of light is always constant? (T or F) 4. Heat can flow from cold to hot (T or F) 5. Sound waves are transverse waves. (T or F) 6. What is the definition of a wave? 7. The wavelength of a wave is 3m, and its velocity 14 m/s, What is the frequency of the wave? 8. Why does an objects temperature not change while it is melting?
1. True: With sound waves, pitch is related to frequency.
2. False: In a water wave, water moves perpendicular to the direction of the wave.
3. True: The speed of light is always constant.
4. False: Heat flows from hot to cold.
5. False: Sound waves are longitudinal waves.
6. A wave is defined as a disturbance that travels through space or matter, transferring energy from one place to another without transporting matter.
7. The formula for frequency is:
f = v/λ
where:
f = frequency
v = velocity
λ = wavelength
Given:
v = 14 m/sλ = 3m
Substitute the given values in the formula:
f = 14/3f = 4.67 Hz
Therefore, the frequency of the wave is 4.67 Hz.
8. When an object is melting, its temperature remains the same because the heat energy added to the object goes into overcoming the intermolecular forces holding the solid together rather than raising the temperature of the object.
Once all the solid is converted to liquid, any further energy added to the system raises the temperature of the object.
This is known as the heat of fusion or melting.
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Suppose a muon produced as a result of a cosmic ray colliding with a nucleus in the upper atmosphere has a velocity v = 0.950c. Suppose it travels at constant velocity and lives 2.20 us as measured by an observer who moves with it (this is the time on the muon's internal clock). It can be shown that it lives for 7.05 us as measured by an Earth-bound observer. (a) How long (in us) would the muon have lived as observed on Earth if its velocity was 0.829c? 3.934e-6 x us (b) How far (in m) would it have traveled as observed on Earth? m (c) What distance in m) is this in the muon's frame? m
a) If the muon's velocity is 0.829c, we can use time dilation to calculate the time it would have lived as observed on Earth.
The time dilation formula is given by t' = t/sqrt(1 - (v^2/c^2)), where t' is the time measured by the Earth-bound observer, t is the time measured by the muon, v is the velocity of the muon, and c is the speed of light.
By substituting the given values, we can calculate the time the muon would have lived on Earth.
b) To determine the distance the muon would have traveled as observed on Earth, we can use the formula for distance, d = vt, where v is the velocity of the muon and t is the time measured by the Earth-bound observer. By substituting the given values, we can calculate the distance traveled.
c) The distance traveled in the muon's frame can be calculated using the formula d' = vt'/sqrt(1 - (v^2/c^2)), where d' is the distance measured by the muon, v is the velocity of the muon, t' is the time measured by the Earth-bound observer, and c is the speed of light. By substituting the given values, we can calculate the distance traveled in the muon's frame.
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1,
If, after you complete Parts 1 and 2 of this lab, you have this Data:
Launch Height: y = 117 cm
Horizontal Launch Velocity: v = 455 cm/s.
How far, x, does the ball travel?
Give your answer in cm to 3 significant figures (no decimal places)
The ball travels approximately 569 cm horizontally.
How to find how the ball travelsTo find the horizontal distance traveled by the ball, we can use the horizontal launch velocity and the time of flight of the ball. However, since the time of flight is not given, we need additional information to determine the horizontal distance accurately.
If we assume that the ball is launched horizontally and neglect any air resistance, we can use the following kinematic equation to find the time of flight:
[tex]\[ y = \frac{1}{2} g t^2 \][/tex]
Where:
- \( y \) is the launch height (117 cm)
- \( g \) is the acceleration due to gravity (approximately 980 cm/s^2)
- \( t \) is the time of flight
Solving for \( t \) in the above equation, we have:
[tex]\[ t = \sqrt{\frac{2y}{g}} \][/tex]
Substituting the given values:
[tex]\[ t = \sqrt{\frac{2 \times 117}{980}} \][/tex]
Now, we can find the horizontal distance traveled by the ball using the formula:
[tex]\[ x = v \cdot t \][/tex]
Substituting the given values:
[tex]\[ x = 455 \times \sqrt{\frac{2 \times 117}{980}} \][/tex]
Calculating the value of \( x \):
[tex]\[ x \approx 569 \, \text{cm} \][/tex]
Therefore, the ball travels approximately 569 cm horizontally.
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