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3 years ago

11

Rest and motion are relative terms why?

1 answer:

krok68 [10]3 years ago

3 0

Answer:

Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body.

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A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th

zheka24 [161]

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Learn more on momentum here: brainly.com/question/402617

5 0

3 years ago

a ball is dropped from rest at a height of 89m above the ground. (a)what is it's speed just before it hits the ground? (b) how l

kykrilka [37]

Answer:

(a) 41.75m/s

(b) 4.26s

Explanation:

Let:

Distance, D = 89m

Gravity, $g$ = 9.8 m/ $s^{2}$

Initial Velocity, $u$ = 0m/s

Final Velocity, $v$ = ?

Time Taken, $t$ = ?

With the distance formula, which is

D = $ut$ + $\frac{1}{2} gt^2$

and by substituting what we already know, we have:

89 = $\frac{1}{2}$ ×9.8× $t^{2}$

With the equation above, we can solve for $t$ :

$t=\sqrt{\frac{89(2)}{9.8}} \\t=\sqrt{\frac{178}{9.8} } \\t=\sqrt{18.16} \\t=4.26 seconds$

Now that we have solved $t$ , we can use the following velocity formula to solve for $v$ :

$v = u + at$ , where $a$ is also equals to $g$ , so we have

$v = u + gt$

By substituting $u = 0$ , $g = 9.8$ , and $t = 4.26$ ,

We have:

$v = 0 + 9.8(4.26)\\v = 41.75m/s$

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3 years ago

Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo

luda_lava [24]

Answer:

?

Explanation:

8 0

2 years ago

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of

Nesterboy [21]

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

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2 years ago

what kind of weather is associated with high pressure?How does density,humidity, and air motion compare to that low pressure sys

frutty [35]

Usually nice weather , i dont know the answer to the second part

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3 years ago