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kari74 [83]
4 years ago
8

What cause earthquakes?

Physics
2 answers:
KengaRu [80]4 years ago
6 0
Question; what causes earthquakes? 
Answer; what causes earthquakes are mainly when rocks are underground and then suddenly break along a fault. 
Brums [2.3K]4 years ago
3 0
When rock underground suddenly breaks along a fault. This sudden release of energy causes seismic waves that make the ground shake. hope this helps!
You might be interested in
1.A particle is moving up an inclined plane. Its velocity changes from 15m/s to 10m/s in two seconds. What is its acceleration?
FrozenT [24]

\huge{ \underline{ \boxed{ \bf{ \blue{Solution:}}}}}

<h3><u>Provided</u><u>:</u><u>-</u></h3>
  • Initial velocity = 15 m/s
  • Final velocity = 10 m/s
  • Time taken = 2 s

<h3><u>To FinD:-</u></h3>
  • Accleration of the particle....?

<h3><u>How</u><u> </u><u>to</u><u> </u><u>solve</u><u>?</u></h3>

We will solve the above Question by using equations of motion that are:-

  • v = u + at
  • s = ut + 1/2 at²
  • v² = u² + 2as

Here,

  • v = Final velocity
  • u = Initial velocity
  • a = acceleration
  • t = time taken
  • s = distance travelled

<h3><u>Work</u><u> </u><u>out</u><u>:</u></h3>

By using first equation of motion,

⇛ v = u + at

⇛ 10 = 15 + a(2)

⇛ -5 = 2a

Flipping it,

⇛ 2a = -5

⇛ a = -2.5 m/s² [ANSWER]

❍ Acclearation is negative because final velocity is less than Initial velocity.

<u>━━━━━━━━━━━━━━━━━━━━</u>

5 0
4 years ago
A rubber ball with a mass of 0.145 kg is dropped from rest. From what height (in m) was the ball dropped, if the magnitude of th
Elena-2011 [213]

Answer:

1.55 m

Explanation:

Momentum: This can be defined as the product of  mass of a body and it velocity. the S.I unit of momentum is kgm/s.

Mathematically,

Momentum can be represented as,

M = mv................................. Equation 1

Where m = mass of the body, v = velocity of the body, M = momentum.

Making v the subject of the equation,

v = M/m........................................... Equation 2

Given: M = 0.80 kg.m/s, m = 0.145 kg.

Substituting into equation 2,

v = 0.8/0.145

v = 5.52 m/s.

Using the equation of motion,

v² = u² + 2gs ....................... Equation 3.

Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.

Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².

Substituting into equation 2

5.52² = 0² + 2(9.81)s

30.47 = 19.62s

s = 30.47/19.62

s = 1.55 m.

Thus the ball was dropped from a height of 1.55 m

8 0
4 years ago
What is the name of the perceived change in a sound wave’s frequency due to motion between the observer and the sound source?
Scrat [10]

when observer and source moves relative to each other then the frequency received by the observer is different from the real frequency

This apparent change in frequency due to relative motion is known as Doppler's effect.

Here we know that

f_{app} = f_o\frac{v\pm v_o}{v \pm v_s}

here we know that

f_o = real frequency

v = speed of sound

v_o = speed of observer

v_s = speed of source

so this is known as Doppler's Effect

6 0
4 years ago
Read 2 more answers
What are six reasons for having goals
ohaa [14]

Answer:

Without a goal, your efforts can become disjointed and often confusing

Helps you measure progress

Helps you stay motivated

Helps you beat procrastination

You Achieve More

Helps you determine what is important

Explanation:

Hope this helps!!!!!!!!!!!

Forever friend and helper,

Cammie:)

3 0
3 years ago
A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no
frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

4 0
4 years ago
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