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3 years ago

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A ball is dropped from the top of an eleven-story building to a balcony on the ninth floor. In which case is the change in the p

otential energy associated with the motion of the ball the greatest

1 answer:

lozanna [386]3 years ago

5 0

Answer:

at the top of the 9 story building i think

Explanation:

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Whenever an object is moving at a constant rate, the value(s) that equal zero is (are):

dolphi86 [110]

Answer:

d) all of the above

Explanation:

the speed, velocity, and acceleration all changes to 0 due to no motion.

5 0

3 years ago

Read 2 more answers

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

3 years ago

A 45.2 kg softball player slides across dirt with Uk=0.340. What is her acceleration?

alisha [4.7K]

Answer:

the ball didnt hit my face so

Explanation:

8 0

3 years ago

Photoelectrons with a maximum speed of 8.00 • 106 m/sec are ejected froma surface in the presence of light with a frequency of 6

Andru [333]

The kinetic energy is given by:

$K=\frac{1}{2}mv^2$

We know the mass and the maximum speed, plugging their values in the expression above we have:

$\begin{gathered} K=\frac{1}{2}(9.1\times10^{-31})(8\times10^6)^2 \\ K=2.91\times10^{-17}\text{ J} \end{gathered}$

Therefore, the answer is d.

5 0

1 year ago

Real springs have mass. How will the true period andfrequency

Ad libitum [116K]

Explanation:

An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.

Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.

4 0

3 years ago