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Andrei [34K]
3 years ago
13

What kind of waves are present during an earthquake? ...

Physics
1 answer:
Brrunno [24]3 years ago
6 0

Answer:

1. The two main types of waves are body waves and surface waves. Body waves can travel through the earth's inner layers, but surface waves can only move along the surface of the planet like ripples on water. Earthquakes radiate seismic energy as both body and surface waves.

2. potential energy

3. Newton's second law of motion is F = ma, or force is equal to mass times acceleration.

4. Refraction is the bending of light

5. Density uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). Density is defined as mass per unit volume.

Explanation:

You might be interested in
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
1. How much energy is needed just to melt 0.56kg of ice at 0◦ C?
ivann1987 [24]

Answer:

This will require 266.9 of heat energy.

Explanation:

To calculate the energy required to raise the temperature of any given substance, here's what you require:The mass of the material, m The temperature change that occurs,  ΔT The specific heat capacity of the material,

c

(which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.

Here is a source of values of

c for different substances:

Once you have all that, this is the equation:

Q=m×c×ΔT(Q is usually used to symbolize that heat required in a case like this.)For water, the value of c is 4.186g°C So, Q=750×4.186×85=266=858=266.858

5 0
2 years ago
An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down
soldier1979 [14.2K]

Answer:

1.9\times 10^{-4}

1.2\times 10^{-4}\ m

Explanation:

r = Radius = 2.7 cm

F = Force = 3.2\times 10^4\ N

A = Area = \pi r^2

\sigma = Stress = \frac{F}{A}

E = Young's modulus = 7\times 10^{10}\ Pa

\epsilon = Strain

L_0 = Original length = 67 cm

\Delta L = Change in length

Young's modulus is given by

E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}

Strain is 1.9\times 10^{-4}

Strain is given by

\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m

The cylinder height decreases by 1.2\times 10^{-4}\ m

3 0
3 years ago
If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52
castortr0y [4]

Answer:

A.\hspace{3}n=1.52

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

n=\frac{c}{v}   (1)

Where:

c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\

v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}

Finally, replacing the data in (1):

n=\frac{300,000}{196339.48} =1.527965746\approx1.52

3 0
3 years ago
Alex places 2 cubes side-by-side on a ramp made of wood. Cube #1 is ice and Cube #2 is wood
zmey [24]

Answer:

Explanation:

The sandpaper block did not move because the forces of friction and gravity were balanced.

6 0
3 years ago
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