Help me, please !! lolll

Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet of

Harrizon [31]

Answer:

a) 6.57 m/s

b) 53.75 J

c) 6.37 m/s

d) -98.297 J

Explanation:

mass of player = $m_{p}$ = 117.5 kg

speed of player = $v_{p}$ = 6.5 m/s

mass of ball = $m_{b}$ = 0.43 kg

velocity of ball = $v_{b}$ = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.5^{2}$ = 2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

$m_{p}$ $v_{p}$ + $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = 6.57 m/s

b) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.57^{2}$ = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained

c) if they travel in opposite direction, equation becomes

$m_{p}$ $v_{p}$ - $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 = 6.37 m/s

d) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.37^{2}$ = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is 98.297 J lost

3 0

4 years ago

A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne

mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0

4 years ago

If the initial velocity of an object was -2 meters per second

Shalnov [3]

A :-) for this question , we should apply
a = v - u by t
Given - u = -2 m/s
v = -10 m/s
t = 16 sec
Solution -
a = v - u by t
a = -10 - -2 by 16
a = -12 by 16
( cut 12 and 16 because 2 x 6 = 12 and
2 x 8 = 16 )
( cut 6 and 8 because 2 x 3 = 6 and
2 x 4 = 8 )
a = 3 by 4
a = 0.75 m/s^2

.:. The acceleration is 0.75 m/s^2.

3 0

3 years ago

a 7.5 kg bowling ball is in front of a large, compressed spring. The spring has a constant of 500 N/m, and 225N of force was req

Vikentia [17]

F=kx
X=F/k=225/500=.45 meters

6 0

3 years ago

The charge an electron has spinning around a nucleus. (What is it?)

Evgen [1.6K]

Answer:

Negative charge

Explanation:

Electron is an negative charge that spins around the nucleus.

4 0

4 years ago

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