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Svetlanka [38]
2 years ago
15

You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley

ball?
Physics
2 answers:
FinnZ [79.3K]2 years ago
8 0

awnser :

Explanation:

Lena [83]2 years ago
3 0

Answer:

I believe the answer is KE = 945 J

Explanation:

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In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t
muminat
<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So,  m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is  m/s2

Acceleration  m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation  ,

m/s

Therefore, the final velocity or jumping speed is  m/s

Explanation:

3 0
3 years ago
Read 2 more answers
Help ASAP plssssssssss
garik1379 [7]
C. 32400 because 540 * 60 = 32400
5 0
3 years ago
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca
4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0
3 years ago
A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
 We have then:
 1 = jug
 2 = skateboard + boy
 By conservation of the linear amount of movement:
 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
 -24.96 = (-0.65) M2
 M2 = (-24.96) / (- 0.65) = 38.4 kg
 Then, the child's mass is:
 M2 = Mskateboard + Mb
 Clearing:
 Mb = M2-Mskateboard
 Mb = 38.4 - 1.9
 Mb = 36.5 Kg
 answer:
 the boy's mass is 36.5 Kg
4 0
3 years ago
Which kind of force do you exert on an object when you pull it toward you?
jarptica [38.1K]

By definition we have to:

Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

Answer:

you exert an Applied Force on an object when you pull it towards you

A. Applied Force

4 0
3 years ago
Read 2 more answers
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