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Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on

VikaD [51]

Answer:

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect

system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is

reversed. As a result, the velocity of the insect changes to a great amount. On the other hand,

the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect

suffers a greater change in momentum as compared to the car is correct. The momentum of

the insect after collision becomes very high because the car is moving at a high speed.

Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to

the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces

caused by the Newton’s action-reaction law. But, he made an incorrect statement as the

system suffers a change in momentum because the momentum before the collision is equal to

the momentum after the collision.

4 0

3 years ago

An instrument is defective for all of the following conditions EXCEPT:_______

NARA [144]

Answer:

I believe its C

6 0

4 years ago

If a quarterback applies a 100N force on a football for a distance of .5m, how much work does he do on the ball

irga5000 [103]

Explanation:

Work done = F * d = 100N * 0.5m = 50J.

3 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

3 years ago

What happens to atomic numbers as you follow a period across the periodic table

geniusboy [140]

Answer:

As you move across a period, the atomic mass increases because the atomic number also increases

Explanation:

When the atomic number increases, this means that there are more protons and neutrons that add to the atomic mass of an atom.

8 0

3 years ago