What are some examples of pressure

The acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. Th

Lorico [155]

1/3 the weight than it is on earth, duh

3 0

2 years ago

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Lowest frequency of wave

Harlamova29_29 [7]

The answer would be a radio wave

4 0

2 years ago

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Help Meeee Please!!!!! This was due so long ago... DO NOT REPORT, THERE IS NO REASON TO REPORT! Assignment: Weathering in Action

zaharov [31]

Answer: Chemical

Explanation: because it is changeing its chemical structure if its wrong sorry .

3 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago

If Galileo drops a cannon ball from the 60 meter high) Leaning Tower of Pisa, how fast will it be moving when it hits the ground

viva [34]

Answer:

When the ball hits the ground, the velocity will be -34 m/s.

Explanation:

The height and velocity of the ball at any time can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity. (-9.8 m/s² considering the upward direction as positive).

v = velocity at time "t".

If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:

y = y0 + v0 · t + 1/2 · g · t²

0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 = 60 m - 4.9 m/s² · t²

-60 m / -4.9 m/s² = t²

t = 3.5 s

Now, with this time, we can calculate the velocity of the ball when it reaches the ground:

v = v0 + g · t

v = 0 m/s - 9.8 m/s² · 3.5 s

v = -34 m/s

When the ball hits the ground, the velocity will be -34 m/s.

5 0

3 years ago