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BARSIC [14]
3 years ago
8

A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l

eft. The bowling ball barely slows down. What is the approximate velocity of the beach ball after the collision?
Physics
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

v'_2=3V

Explanation:

From the question we are told that:

Bowling ball Speed v_1=5 m/s

Beach ball Speed v_2=2 m/s

Let The Mass be equal i.e

 M_1=M_2

Therefore

Generally the equation for Velocity of beach ball after collision v'_2 is mathematically given by

Since Velocity is Vector Quantity

Therefore

 v'_2=v_1-v_2

 v'_2=5-2

 v'_2=3V

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Answer:

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A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the
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EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

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When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *
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You accidentally drop an eraser out of the window of an apartment 15 m above the ground
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2 years ago
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
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