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3 years ago

12

Which type of earthquake waves are responsible for most of the damage caused by an earthquake? Why?

2 answers:

larisa [96]3 years ago

7 0

Answer:

surface waves

Explanation:

they have the strongest vibration and can cause the most damage

den301095 [7]3 years ago

5 0

Answer:

Surface waves travel along the surface. There are two types of body waves: P-waves travel fastest and through solids, liquids, and gases; S-waves only travel through solids. Surface waves are the slowest, but they do the most damage in an earthquake.

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The kinetic energy of an object depends on which of the following?

aleksandr82 [10.1K]

B. the object's speed and mass

Looking at the kinetic energy formula:

1/2mv^2 = Ek

It can be seen that the only two factors that affect the magnitude of the kinetic energy are mass and the object speed. Kinetic energy is also a scalar quantity so the direction is unnecessary.

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3 years ago

Read 2 more answers

How much of the electro magnetic spectrum is visible to us

Rudiy27

Answer:

The entire rainbow of radiation observable to the human eye only makes up a tiny portion of the electromagnetic spectrum – about 0.0035 percent. This range of wavelengths is known as visible light.

Explanation:

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3 years ago

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

What class lever is a

Alchen [17]

should be 1

Explanation:

because what do you use in class and has a lever

8 0

3 years ago

A(n) ____ like rubber or plastic has high electrical resistance.

blagie [28]

Answer:

insulator

Explanation:

6 0

2 years ago