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3 years ago

Which unit is equivalent to j/s

1 answer:

4 0

To be honest, there's no sure way to answer that, because you haven't defined your terms and we can't be sure of what j or s might be.

Tell you what I'll do:. I'll assume definitions for j and s, and then I'll answer the question that I invented.

Assume that j stands for Joule, the unit of energy. And assume that s stands for 'second', the unit of time.

Then j/s is the rate of transferring energy or doing work.

Its unit is the Watt, equivalent to 1 Joule per second.

In your system of notation, it would be 'w' .

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A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

What is the largest area of specialization in psychology

Sindrei [870]

Answer:clinical psychology

Explanation:

The largest area of specialization in psychology is clinical psychology

5 0

3 years ago

Question If a force of 2N is used to move a box 1 meter, how much work was done to move the box?

Sunny_sXe [5.5K]

Answer:

I assume that the force of 2N is applied along the direction of motion and was applied for the whole 1 meter, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula;

Explanation:

Work = 2N * 1m * cos 0º.

Work = 2N * 1m * 1

Work = 2Nm

Work = 2 joules

7 0

3 years ago

Why is the universe here and why do we exist?

8_murik_8 [283]

Become a physicist and maybe one day you can tell me.

8 0

3 years ago

A square sheet of rubber has sided that are 20 cm long. If you stretch one side so that it is twice as long. What is the new are

Anika [276]

Answer:ignore this it was wrong

Explanation:

3 0

4 years ago