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2 years ago

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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.

1 answer:

Inessa [10]2 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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<span>We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_(f) = p_(1) + p_(2) + p_(3) = 0.
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Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
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2 years ago

Read 2 more answers

A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while

rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

$P_1V_1 =P_2 V_2$

Where,

$P_1$ and $V_1$ are the initial pressure and volume values

$P_2$ and $V_2$ are the final pressure volume values

The final pressure here is the atmosphere, then

$P_2 = 101325 \approx 1*10^5Pa$

$h = 10m$

$\rho_w = 1000kg/m^3$

$V_1 = 3.0L$

Pressure at the water is given by,

$P_1 = P_2 -\rho gh$

$P_1 = 1*10^5 +1000*9.8*10 =198000Pa$

Using Boyle equation we have,

$V_2 = \frac{P_1V_1}{P_2}$

$V_2 = \frac{198000*3*10^5}{10^5}$

$V_2 = 5.9L$

Therefore the volume of the lungs at the surface is 5.9L

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2 years ago

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

$W = q \Delta V$

where

$q=3.0 \mu C = 3.0 \cdot 10^{-6} C$ is the magnitude of the charge

$\Delta V = 6.0 kV = 6000 V$ is the potential difference between point P and infinity

Substituting into the equation, we find

$W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J$

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2 years ago

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Yuki888 [10]

Answer:

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Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

$(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)$

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$0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t$

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El guijarro tiene una rapidez de $5.219\,\frac{m}{s}$ cuando golpea la ventana (The peeble has a speed of $5.219\,\frac{m}{s}$ when it hits the window).

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2 years ago