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erik [133]
4 years ago
9

The magnitude, M, of an earthquake is represented by the equation M=23logEE0 where E is the amount of energy released by the ear

thquake in joules and E0=104.4 is the assigned minimal measure released by an earthquake. In scientific notation rounded to the nearest tenth, what is the amount of energy released by an earthquake with a magnitude of 5.5?
Physics
1 answer:
vfiekz [6]4 years ago
8 0

Answer:

The amount of energy released by the earthquake in joules is 4.4×10^12 J.

Explanation:

As, we have given the magnitude,

M = 2/3 Log E/Eo

where E is the amount of energy released by the earthquake in joules and Eo=10^4.4 is the assigned minimal measure released by an earthquake.

As, the magnitude is given which is 5.5 then put it in the above equation,

5.5 = 2/3 Log E/Eo

Log E/Eo = 5.5×3 / 2

Log E/Eo = 8.25

Now, we will find the amount of energy released by an earthquake:

Log E/Eo = 8.25

Taking antilog,

E/Eo = 10^8.25

But Eo is given which is Eo = 10^4.4

E/(10^4.4) = 10^8.25

E = 10^8.25 × 10^4.4

E = 4.4×10^12 J.

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Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

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x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

w = \sqrt{\frac{k}{m}}

w = \sqrt{\frac{250}{0.5}}

w = 22.4 rad/s

f = frequency

Angular frequency is also given as

w = 2 π f

22.4 = 2 (3.14) f

f  = 3.6 Hz

T = Time period

Time period is given as

T = \frac{1}{f}

T = \frac{1}{3.6}

T = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

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4 years ago
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Complete question is;

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Explanation:

We want to find the one that is a qualitative observation;

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