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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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An athlete can twirl a baton so that it reaches an angular velocity of 12 rad/s from rest in 0.5 s. If the length of a uniform b
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Answer:

The torque needed is 46.08 Nm

Explanation:

Given;

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time of motion, t = 0.5 s

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mass of the baton, 0.5 kg

The torque needed to  reach the angular velocity is given by;

τ = F x r

where;

F is the centripetal force of the baton

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F_c =ma_c= \frac{mv^2}{r} = m\omega^2 r

Torque is given by;

\tau = F_c *r\\\\\tau = m \omega^2 r *r\\\\\tau = m \omega^2 r^2 \\\\\tau = (0.5)(12)^2(0.8)^2\\\\\tau = 46.08 \ N m

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A 2.3 kg , 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnt
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Answer

ω2=82.1 rpm

Explanation:

given                                    required              

m1=2.3 kg                           ω2=?              

m2+m3=0.5kg

r1=10 cm

ω1=100 rpm

solution

Using the application of conservation of angular momentum we can solve as follows/

L before collision= L after collision

m1r²ω1=(m1+m2+m3)r²ω2

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2.3 rpm=0.028×ω2

ω2=82.1 rpm

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4 years ago
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