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3 years ago

Earthquakes typically occur

Physics

1 answer:

ICE Princess25 [194]3 years ago

3 0

Answer:

<h3><u><em>Earthquakes are usually caused when rock underground suddenly breaks along a fault. This sudden release of energy causes the seismic waves that make the ground shake. When two blocks of rock or two plates are rubbing against each other, they stick a little. ... When the rocks break, the earthquake occurs.</em></u></h3>

Explanation:

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What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure

Ivahew [28]

<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

Pressure = Force/Area

★ Putting the values in the above formula,we get:

$\sf \rightarrow \: pressure = \dfrac{4000}{50}$

$\sf \rightarrow pressure = 80 \: N {m}^{ - 2}$

7 0

3 years ago

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

3 0

3 years ago

A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0

4 years ago

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

7 0

3 years ago

Does exists the friction force in space?If yes,tell an full example

VashaNatasha [74]

Yes, friction does exist in space. Friction has nothing to do with the earth's atmosphere. It exists everywhere in the universe. <span />

5 0

3 years ago