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3 years ago

What equation do we use to solve?

Physics

1 answer:

aev [14]3 years ago

4 0

V(y) is the same in all cases so time taken for all cannonballs to go up and down is the same.

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An aurora occurs when _____.

drek231 [11]

The correct answer is A. Charged particles from the sun exite the atmosphere near the poles to create auroras.

6 0

3 years ago

Read 2 more answers

The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t

REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

From the question we are told that

The initial Volume is $Vi = 0.160 L$

The final volume is $V_f = 0.510L$

The external pressure is $P = 5.00 \ atm$

Generally the change in volume is

$\Delta V = V_f - V_i$

Substituting values we have

$\Delta V = 0.510 -0.160$

$= 0.350L$

Generally workdone is mathematically represented as

$W = -P \Delta V$

W is negative because the working is done on the environment by the system which is indicated by volume increase

Substituting values

$W = - 5* 0.350$

$= -1.75 \ L \ \cdot atm$

Now $1 \ L \cdot atm = 101.3J$

Therefore $W = -1.75* 101.3$

$= -177.275J$

7 0

3 years ago

Read 2 more answers

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$