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Gre4nikov [31]
2 years ago
13

What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of

friction?
volcano


hot spot


earthquake


trench
Physics
2 answers:
Dahasolnce [82]2 years ago
7 0
What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of friction?


earthquake


Arlecino [84]2 years ago
4 0
It has to be Earthquake
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A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has
Nookie1986 [14]

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block \rho_{b} =9.50\times10^2\ kg/m^3

Density of fluid \rho_{t} =1.30\times10^3\ kg/m^3

We need to calculate the depth

Using balance equation

mg=\rho g V....(I)

We know that,

The density is

\rho=\dfrac{m}{V}

m=\rh0\times V

Put the value of m in equation (I)

\rho_{b}\times V\times g=\rho_{t}\times g\times V

\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h

h=\dfrac{\rho_{b}H}{\rho_{t}}

Put the value into the formula

h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}

h= 5.85\ cm

h=0.1919291\ ft

We need to calculate the acceleration

Using formula of net force

F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V

ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V

\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V

a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g....(II)

Put the value in the equation (II)

a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8

a=3.61\ m/s^2

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

4 0
3 years ago
Explain two ways you could reduce the friction between two surfaces
abruzzese [7]
<span>A lubricant such as oil, grease, graphite powder  can reduce the friction between two surfaces. Or using metal balls to space them and reduce the contact surface area as used in ball bearings.</span>
3 0
3 years ago
How fast does a 2 MeV fission neutron travel through a reactor core?
Artemon [7]

Answer:

The answer is 1.956 \times 10^7\ m/s

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy E = mc^2, so the definition of energy in this case is

E = \frac{1}{2}m v^2.

From the last equation,

v= \sqrt{2E/m}

where

E = 2 MeV = 3.204 \times 10^{-13} J

and the mass of the neutron is

m = 1.675\times 10^{-27}\ Kg.

Then

v = 1.956 \times 10^7\ m/s

the equivalent of 0.065 the speed of light.

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If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di
xenn [34]

The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered.  But it can never be greater than the distance you covered.

This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.

The straight line is always the shortest path between two points.

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If two identical trees are cut down, one with a hand saw, and one with an electric saw...
nataly862011 [7]
The hand saw would involve more work because it takes more time and effort. 
4 0
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