Ask question

Ask question

All categories

4 years ago

8

A ball took 0.45s to hit the ground 0.72m from the table. What was the horizontal velocity of the ball as it rolled off the tabl

e?

1 answer:

dusya [7]4 years ago

8 0

the ball hits the ground a very high velocity

You might be interested in

If frequency is kept constant, how are the velocity and wavelength of a wave related?

Elan Coil [88]

<h3><u>Answer;</u></h3>

Velocity and wavelength are directly proportional when frequency is kept constant.

<h3><u>Explanation;</u></h3>

<em><u>Frequency of a wave is the number of complete oscillations made by a given wave in one second. </u></em>
<em><u>Wavelength on the other hand, is the distance between two successful crests or troughs in a transverse wave or two successful rarefactions or compressions in a longitudinal waves.</u></em>
<em><u>The speed of a wave is given by the product of the frequency of a wave and the wavelength.</u></em>
<em><u>Speed = Frequency × wavelength, </u></em>
<em><u>Therefore, if frequency is kept constant, then the speed of a wave is directly proportional to the wavelength, such that an increase in wavelength increases the speed of the wave and vice versa.</u></em>

3 0

3 years ago

Read 2 more answers

Which part of the larynx is the opening between the vocal folds where vibration occurs?

Lynna [10]

Glottis I believe is the answer.

5 0

3 years ago

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

How remove local action and polarization

slava [35]

Answer:

Local action is removed by amalgamating zinc rod.

Explanation:

This prevents chemical reaction to occur because impurities in the zinc cannot get into contact with an electrolyte.

8 0

4 years ago

For Valentine’s Day, Sally received a helium-filled balloon at a party. On returning home she accidentally left the balloon in t

Softa [21]

Answer:

If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.

As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.

8 0

3 years ago