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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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• What value is used to identify each element in the periodic table?​
bearhunter [10]

Answer:

protons

Explanation:

5 0
3 years ago
I need the answer to this question what has the student plotted on the vertical axis?
expeople1 [14]

Answer:

The correct option is D

Explanation:

In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.

7 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
In order to increase the amount of work done, we need to:
Nadusha1986 [10]
Option (D) is the correct one.


In order to increase the amount of work done, we need to increase the force applied to the object.
8 0
3 years ago
Heat transfer between two substances is affected by specific heat and the
IgorLugansk [536]

The correct answer is D. Amount of time and area of physical contact between the substances.

Explanation:

Heat transfer refers to the flow of thermal energy or heat between two or more objects. This process involves multiple factors and implies heat from the hottest object goes to the coldest one until there is an equilibrium. To begin, heat transfer depends on the amount of thermal energy in the objects because objects must have a different amount of thermal energy for heat to flow.

Besides this, the amount of energy that flows depends on the time and the contact between the substances of objects. Indeed, objects need to be in contact or close to each other for heat to transfer, and the time needs to be enough for the process to occur. For example, if you place a pot over the fire just for a few seconds it is likely the heat transferred is minimal, which does not occur if you leave the pot more time. At the same time if the pot is in close contact with fire more heat will be transferred.-

6 0
3 years ago
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