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Hitman42 [59]
3 years ago
10

A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m

/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. Answer in units of m/s.
Physics
1 answer:
Inessa [10]3 years ago
4 0

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

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Answer: a) Cnew=Cinitial ; b) λouter new=  2*λ outer initial

Explanation: In order to explain this question we have to take into account the expression of teh cylinder capacitor given by:

C/L= (2*π*εo)/ln (b/a)=  where b and a are the outer and inner radius, respectively. L is the length of the capacitor.

As you can se this formule depents  of geometrical characateristics  of the capacitor.

The capacitance is the same after change the densities of charge.

On the other hand,

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The new potential is 2 times the initial one so

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Also we know that

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using this formule and considering  that V new is doubled then the charge per one meter length,  is also doubled .

This is as follow:

Vnew=  λnew*L/C=  

λnew =  (2*Vinitial)* C/L= 2  (λ initial)

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What is sap?
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Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

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The emf of the inductor is given by;

V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

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