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Lunna [17]
2 years ago
15

A particle was moving in a straight line at 172.8 km/hr. If it decelerated over 120 meters to come to rest, find the time taken

to cover this distance.​
Physics
1 answer:
Bezzdna [24]2 years ago
8 0

Answer:

v=s/t

s=vt

t=s/v

t=(120×10‐³)/172.8

(the distance meters has been changed to kilometres)

t=1/1440 hrs

Given ,

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Transverse waves are generally stronger than longitudinal waves. True or False ?
kondaur [170]
False
If all other factors, such as medium, are kept the same, longitudinal waves tend to be stronger.
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3 years ago
What do you understand by ?efficiency of a machine is 70%​
tatuchka [14]

Explanation:

The efficiency of a machine is 70% means 30% of the applied force is wasted in overcoming friction and 70% of the applied force is used to do the work

8 0
3 years ago
The magnitude of the angular momentum of the two-satellite system is best represented by
zloy xaker [14]

The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

<h3>What is angular momentum.?</h3>

The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.

The magnitude of the angular momentum of the two-satellite system is best represented as;

L=∑mvr

L=m₁v₁r₁-m₂v₂r₂

Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

To learn more about the angular momentum, refer to the link;

brainly.com/question/15104254

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3 0
2 years ago
An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin
RUDIKE [14]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

4 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
2 years ago
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