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kvv77 [185]
2 years ago
4

A ball is tossed straight up and later returns to the point from which it was launched. If the ball is subject to air resistance

as well as gravity, which of the following statements is correct?
a. The net work done by gravity on the ball during its flight is greater than zero.
b. The speed at which the ball returns to the point of launch is less than its speed when it was initially launched.
c. The net work done by air resistance on the ball during its flight is zero.
d. The force of air resistance is directed down-ward both when the ball is rising and when it is falling.
Physics
1 answer:
Amiraneli [1.4K]2 years ago
4 0

Answer:

c. The net work done by air resistance on the ball during its flight is zero.

Explanation:

  • When a ball is tossed up then it experiences a force of gravity and a force of air-drag in the direction opposite to the motion. When the ball goes up it acts in the downward direction and when the ball goes down it acts upward.
  • The speed and time is symmetric in both the cases whether the ball goes up or it comes down after being tossed straight up.
  • The work done by the gravity is zero because ultimately when the ball comes down its displacement is zero and hence work done is zero. Similarly for the air resistance the displacement is zero leading the work to be zero.

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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational
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Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: F = 49.06N

Pebble: F = 29.44N

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: a =9.8m/s^{2}

Pebble:  a =9.8m/s^{2}

Explanation:

The universal law of gravitation is defined as:

F = G\frac{m1m2}{r^{2}}  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em>m = 5.0 Kg<em>:</em>

m1 will be equal to the mass of the Earth m1 = 5.972×10^{24} Kg and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth r = 6371000m.

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}

F = 49.06N

Newton's second law can be used to know the acceleration.

F = ma

a =\frac{F}{m} (2)

a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}

a =9.8m/s^{2}

<em>Case for the pebble </em>m = 3.0 Kg<em>:</em>

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}

F = 29.44N

a =\frac{F}{m}

a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}

a =9.8m/s^{2}

3 0
2 years ago
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PART B)

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