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Aleksandr [31]
3 years ago
13

What type of energy dose an earthquake transfer

Physics
1 answer:
IrinaVladis [17]3 years ago
7 0
Earthquakes release energy in several forms: The energy in seismic waves that cause the ground to shake. Heat energy associated with friction on the fault slip surface. Gravitational potential energy (the energy stored when lifting something off the ground, for example) may change as a result of the earthquake.
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A man on a bicycle of total mass of 10kg has a Velocity of 2 ms.' He Paddles faster for 5 seconds and the velocity Increase to i
sesenic [268]

Answer:

the value of force, F=4.0N

Explanation:

Firstly, recall velocity-time equation

  • v=u+at
  • (4)=(2)+a(5)
  • a=0.4m/s²

Secondly, recall the Newton's 2nd Law

  • <em>F</em><em>=</em><em>ma</em>
  • <em>F</em><em>=</em><em>(</em><em>1</em><em>0</em><em>)</em><em>(</em><em>0</em><em>.</em><em>4</em><em>)</em>
  • <em>F</em><em>=</em><em>4</em><em>.</em><em>0</em><em>N</em>
7 0
1 year ago
C2H4O2 is what ??????
alukav5142 [94]
Did you try looking it up ?
8 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
Read 2 more answers
How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

3 = (40 - 10) / t          Multiply both sides by t

3*t = t(40 - 10)/t        Combine. Cancel t's on the right

3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

6 0
2 years ago
245289457u2895785/+55891235723895718957=
Bingel [31]

<em>why Is there just a random "u" in the middle of the equation </em><em>awa</em><em> </em>

<em> </em><em>t</em><em>h</em><em>e</em><em>question is impossible to do btw </em>

<em>Imao the u in the middle doesn't makes sense </em>

if<em> you edit the question I'll help you w it)</em>

have a good day bro cya

6 0
2 years ago
Read 2 more answers
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