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Lilit
1 month ago
10

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

descend together a distance of 0.150 m.
How much work do your hands do on the volleyball in the process of stopping it?
What is the magnitude of the force (assumed constant) that your hands exert on the volleyball?
Physics
1 answer:
Softa [3K]1 month ago
3 0

Explanation:

The amount of work performed corresponds to the variation in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work can be defined as the product of force and distance.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

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Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the
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Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

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An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water of density 1000 kg/m3. What is the density of
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1 month ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [3030]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

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