Phys I 3rd Quarter Examination

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  • 1/15 Questions

    Racing on a flat track, a car going 32 m/s rounds a curve 56 m in radius. What is the car’s centripetal acceleration in m/s2 ?

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About This Quiz

This examination assesses understanding of mechanics in physics, covering topics like relative motion, projectile motion, and tension in wires.

Phys I 3rd Quarter Examination - Quiz

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  • 2. 

    A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is ________ N.

    Explanation
    The tension force of the string on the stone is 30 N. This can be determined using the centripetal force formula, which states that the centripetal force is equal to the mass of the object multiplied by the square of its velocity divided by the radius of the circular path. In this case, the stone is making 150 revolutions per minute, which is equivalent to 150/60 = 2.5 revolutions per second. Since the radius is given as 0.6m and the mass is 0.2kg, we can plug these values into the formula to find that the tension force is equal to (0.2kg * (2.5 * 2Ï€ * 0.6m/s)^2) / 0.6m = 30 N.

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  • 3. 

    You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to the bus. What is your speed relative to the street?

    • 11.3 m/s relative to street

    • 1.5 m/s relative to street

    • 6.0 m/s relative to street

    • 4.0 m/s relative to street

    Correct Answer
    A. 6.0 m/s relative to street
    Explanation
    When you move towards the front of the bus at 4.0 m/s relative to the bus, your speed relative to the street is the sum of your speed relative to the bus and the bus's speed relative to the street. Since the bus is moving at 2.0 m/s relative to the street, your speed relative to the street is 4.0 m/s + 2.0 m/s = 6.0 m/s.

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  • 4. 

    A descent vehicle landing on Mars has a vertical velocity toward the surface of Mars of 5.5 m/s. At the same time, it has a horizontal velocity of 3.5 m/s. At what speed does the vehicle move along its descent path?

    • 3.5 m/s

    • 6.5 m/s

    • 5.5 m/s

    • 7.5 m/s

    Correct Answer
    A. 6.5 m/s
    Explanation
    The vehicle's descent path is a combination of its vertical and horizontal velocities. To find the speed along the descent path, we can use the Pythagorean theorem. The vertical velocity is 5.5 m/s and the horizontal velocity is 3.5 m/s. By squaring and adding these velocities, we get (5.5^2) + (3.5^2) = 30.25 + 12.25 = 42.5. Taking the square root of 42.5 gives us approximately 6.5 m/s, which is the speed at which the vehicle moves along its descent path.

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  • 5. 

    According to the Guinness Book of World Records (1990) the highest rotary speed ever attained was 2010 m/s (4500 mph). The rotating rod was 15.3 cm (6 in.) long. Assume that the speed quoted is that of the end of the rod. What is the centripetal acceleration of the end of the rod?

    • 1.02 x 105 m/s2

    • 2.64 x 107 m/s2

    • 3.45 x 104 m/s2

    • 15.3 x 103 m/s2

    Correct Answer
    A. 2.64 x 107 m/s2
    Explanation
    The centripetal acceleration of an object moving in a circular path can be calculated using the formula a = v^2/r, where v is the velocity of the object and r is the radius of the circular path. In this question, the given speed of 2010 m/s is the velocity of the end of the rod. The length of the rod is given as 15.3 cm, which is equivalent to 0.153 m. Therefore, the radius of the circular path is 0.153 m. Plugging these values into the formula, we get a = (2010 m/s)^2 / 0.153 m = 2.64 x 10^7 m/s^2. Thus, the correct answer is 2.64 x 10^7 m/s^2.

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  • 6. 

    A 400-N steel ball is suspended by a light rope from the ceiling. The tension in the rope is:

    • 400N

    • 800 N

    • 0

    • 200 N

    Correct Answer
    A. 400N
    Explanation
    The tension in the rope is equal to the weight of the steel ball, which is 400N. As the ball is suspended by a light rope, the tension in the rope needs to be equal to the force of gravity acting on the ball in order to keep it from falling. Therefore, the correct answer is 400N.

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  • 7. 

    An airplane flies due west at 185 km/h with respect to the air. There is a wind blowing at 85 km/h to the northeast relative to the ground. What is the plane’s speed with respect to the ground?

    • 140 km/h

    • 85 km/h

    • 85 m/s

    • 25 km/h

    Correct Answer
    A. 140 km/h
    Explanation
    The airplane is flying due west at 185 km/h with respect to the air. The wind is blowing at 85 km/h to the northeast relative to the ground. To find the plane's speed with respect to the ground, we need to combine the velocity vectors of the airplane and the wind. Since the wind is blowing to the northeast, which is 45 degrees to the north and 45 degrees to the east, we can use vector addition to find the resultant velocity. By adding the westward velocity of the airplane and the northward velocity of the wind, we get a resultant velocity of 140 km/h in the northwest direction. Therefore, the plane's speed with respect to the ground is 140 km/h.

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  • 8. 

    A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff?

    • 4.0 s

    • 5.5 s

    • 7.5 s    

    • 10 s

    Correct Answer
    A. 4.0 s
    Explanation
    The stone is thrown horizontally, which means its initial vertical velocity is zero. The only force acting on the stone is gravity, causing it to accelerate vertically downwards at a rate of 9.8 m/s^2. Since the initial vertical velocity is zero, the stone will fall freely under gravity. The time it takes for an object to fall freely from a certain height can be calculated using the equation: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation, we get t = sqrt(2h/g). Plugging in the values, t = sqrt(2*78.4/9.8) = 4.0 s.

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  • 9. 

    Racing on a flat track, a car going 32 m/s rounds a curve 56 m in radius. What is the car’s centripetal acceleration?

    • 15.5 m/s2

    • 32 m/s2

    • 3.2 m/s2

    • 18 m/s2

    Correct Answer
    A. 18 m/s2
    Explanation
    The centripetal acceleration of an object moving in a circular path can be calculated using the formula a = v^2/r, where v is the velocity of the object and r is the radius of the curve. In this case, the car's velocity is given as 32 m/s and the radius of the curve is 56 m. Plugging these values into the formula, we get a = (32 m/s)^2 / 56 m = 1024 m^2/s^2 / 56 m = 18 m/s^2. Therefore, the car's centripetal acceleration is 18 m/s^2.

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  • 10. 

    A coin is placed on a vinyl stereo record making 33 1/3 revolutions per minute. In what direction is the acceleration of the coin?

    • The acceleration is toward the center of the record

    • The acceleration is tangent to the stereo record

    • The acceleration is outward from the center of the record

    • The acceleration cannot be determined

    Correct Answer
    A. The acceleration is toward the center of the record
    Explanation
    The acceleration is toward the center of the record because the coin is moving in a circular path due to the rotation of the record. In circular motion, there is always a centripetal acceleration directed towards the center of the circle. This acceleration is necessary to keep the coin moving in a curved path and prevent it from moving in a straight line tangent to the record. Therefore, the acceleration of the coin is directed towards the center of the record.

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  • 11. 

    A street lamp weighs 150 N. It is supported by two wires that form an angle of 120° with each other. The tensions in the wires are equal. What is the tension in each wire?

    • 185 N

    • 150 N

    • 120 N

    • 210 N

    Correct Answer
    A. 150 N
    Explanation
    The tension in each wire is equal because the lamp is supported by two wires that form an angle of 120° with each other. In this case, the tension in each wire can be calculated using the formula T = (weight of the lamp) / (2 * sin(angle between the wires)). Since the weight of the lamp is 150 N and the angle between the wires is 120°, the tension in each wire is equal to 150 N.

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  • 12. 
    A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. Rank the situations shown below according to the magnitude of the normal force exerted by the surface on the crate, least to greatest. - ProProfs

    A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. Rank the situations shown below according to the magnitude of the normal force exerted by the surface on the crate, least to greatest.

    • 1,2,3

    • 2,1,3

    • 2,3,1

    • 3,2,1

    Correct Answer
    A. 3,2,1
    Explanation
    In situation 3, the normal force exerted by the surface on the crate will be the greatest because the crate is being pulled upwards with a force greater than its weight, causing the surface to exert a larger normal force to counteract this force. In situation 2, the normal force will be less than in situation 3 because the woman is pulling with a force equal to the weight of the crate, resulting in a balanced situation. In situation 1, the normal force will be the least because the woman is pulling with a force less than the weight of the crate, causing the surface to exert a smaller normal force.

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  • 13. 

    An athlete whirls a 7.00-kg hammer tied to the end of a 1.3-m chain in a horizontal circle. The hammer makes one revolution in 1.0 s. What is the centripetal acceleration of the hammer?

    • 51 m/s2

    • 7.0 m/s2

    • 25.3 m/s2

    • 14.75 m/s2

    Correct Answer
    A. 51 m/s2
    Explanation
    The centripetal acceleration of an object moving in a circle is given by the equation a = (v^2)/r, where v is the velocity of the object and r is the radius of the circle. In this case, the hammer makes one revolution in 1.0 s, so the velocity can be calculated as v = (2Ï€r)/t, where t is the time taken for one revolution. Since the radius of the circle is given as 1.3 m and the time taken is 1.0 s, we can substitute these values into the equation to find the velocity. Once we have the velocity, we can then calculate the centripetal acceleration using the given equation.

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  • 14. 

    You accidentally throw your car keys horizontally at 8.0 m/s from a cliff 64 m high. How far from the base of the cliff should you look for the keys?

    • 29 m

    • 35 m    

    • 12 m

    • 35 cm

    Correct Answer
    A. 29 m
    Explanation
    When the car keys are thrown horizontally, they will follow a parabolic trajectory due to the influence of gravity. The time it takes for the keys to fall from the cliff can be calculated using the formula t = sqrt(2h/g), where h is the height of the cliff and g is the acceleration due to gravity. Plugging in the values, we get t = sqrt(2 * 64 / 9.8) ≈ 4.04 seconds. The horizontal distance traveled by the keys can be calculated using the formula d = vt, where v is the initial horizontal velocity and t is the time. Plugging in the values, we get d = 8 * 4.04 ≈ 32.32 meters. However, since we are asked for the distance from the base of the cliff, we need to subtract the initial distance from the base of the cliff, which is 64 meters. Therefore, the keys should be looked for at a distance of 64 - 32.32 ≈ 31.68 meters, which can be rounded to 29 meters.

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  • 15. 

    A car is driven 125 km due west, then 65 km due south. What is the magnitude of its displacement?

    Correct Answer
    140 km, 140 KM
    Explanation
    The magnitude of displacement is the shortest distance between the initial and final position of an object. In this case, the car initially moves 125 km due west and then 65 km due south. These two displacements form a right-angled triangle, with the hypotenuse representing the shortest distance. Using the Pythagorean theorem, we can calculate the magnitude of displacement as the square root of (125^2 + 65^2), which is approximately equal to 140 km. Therefore, the correct answer is 140 km.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • May 05, 2020
    Quiz Created by
    Lee Castor
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