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ElenaW [278]
4 years ago
11

What would be the magnitude of an earthquake 100 km away that produced 100 mm of amplitude

Physics
2 answers:
Masteriza [31]4 years ago
6 0

Answer: 100000000

Explanation:

erastovalidia [21]4 years ago
3 0
100000000 mark me brainliest and i will thanks you and rate you5

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6. A girl pushes her little brother on his sled with a force of 300 N for 750 m.
Oduvanchick [21]

w=225000 J  

E=225000 J  

P=9000 W  

P=22500 W

Ew=Fd=E  

P=E/t

5 0
3 years ago
Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

Answer:

Explanation:

We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.

By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

KE₁=Kinetic energy at location 1

U₁ =Potential energy at location 1

KE₂=Kinetic energy at location 2

U₂=Potential energy at location 2

Therefore, Raymond is thinking in a right way.

7 0
3 years ago
A roller coaster track is 3000 meters long. It takes 100 seconds to travel once around the roller coaster. What is the average s
Svetllana [295]
All you have to do is divide 3000 by 100, Its 30
6 0
3 years ago
Read 2 more answers
Earlier we discussed the concept of isostasy, where lower density rocks rise higher than higher density rocks. How is the variat
Elena L [17]

The variation of water depth at spreading centers (ridges) controlled by isostasy  as convective cooling cools the rocks much more effectively the than heat conduction.

<h3>What is convective heat transfer?</h3>

When heat transfer takes place between the two fluids in direct or indirect contact.

The lithosphere cools when it moves away from the ridge axis by sea floor spreading. The cooler rocks have low density, so the sea floor gets deeper as the lithosphere gets more dense.

Thus, the convective cooling cools the rocks much more effectively the than heat conduction.

Learn more about convective heat transfer

brainly.com/question/10219972

#SPJ1

4 0
2 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
marta [7]
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
6 0
3 years ago
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