A car is moving along the street with acceleration of 0.500 m/s^2. A 200m-long t

Question

A car is moving along the street with acceleration of 0.500 m/s^2. A 200m-long train is moving at 10.0 m/s, parallel to the street. The speed of the car is 170 m/s when it gets the tail of the train. How long will it take for the car to go from the tail to the front of the train? A50.0-kg parachuts falls from a helicopter at 200 m over the earth. Calculate his speed when he gets the earth if the total work done by air resistance is -1.75x10^3 J. A satellite is in a circular orbit around the center of the earth, with an orbital radius this is two times the radius R of the Earth. Determine the period of the circular motion of the satellite. A 75-cm guitar string is fixed at both ends. In the frequency range between 1.0 and 2.0 kHz, the string is found to resonate only at frequencies 12, 15, and 1.8 kHz. What is the speed of travelling on this string. An ideal gas absorbs 600 J heat and expands isothermally. Calculate the exchange in internal energy of the gas during the process. A car wheel with a radius of 50 cm has a white spot in its border. The car moves without sliding with a velocity of 80 km/h. Calculate how often the spot touches the street. A football is kicked at ground level with a speed of 25.0 m/s at an angle of 3.0 degree to the horizontal. What is its velocity at the highest point? Ignore air friction. A travelling wave can be described by the wave function D = 0.01 sin (4.2x + 2100 t), where D and x are in meters and t is in seconds. Determine the wave velocity. Calculate the change in entropy of 250 g of water when it is heated from 10 degreeC to 20 degreeC. You can make an estimate by using an average temperature of 15 degreeC during the process. If 9.00 L of hydrogen gas initially at 17.0 degreeC and 1.00 atm is compressed to 3.00 L and heated to 97.0 degreeC. What will be the new pressure?

Explanation / Answer

1. a = 0.500 m/s^2

when car is at tail,
relative velocity of car wrt car, vi = 17 - 10 = 7 m/s

a = 0.5 - 0 = 0.5 m/s^2

d = 200m

Applying,

d = vi*t + a*t^2/2

200 = 7t + 0.5t^2/2

0.25 t^2 + 7t - 200 = 0

t = 17.56 sec

2. Work done by gravity + work done by air = change in kE

m g h + Wr = m v^2 /2 - 0

(90 * 9.8 * 200 ) + (-1.75 * 10^5) = 90 v^2 /2

v = 5.58 m/s

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