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Gennadij [26K]
3 years ago
5

Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm

per year. (a) What distance does it move in 71 seconds at this speed?
m
(b) What is its speed in miles per million years?
mi/My
Physics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

6=\frac{6}{365.25\times 24\times 60\times 60}

So in one second it will move

\frac{6}{365.25\times 24\times 60\times 60}

In 71 seconds

71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My

Speed of the tectonic plate is 37.278 mi/My

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Explanation:

(a) In order to find the work done by the applied force you use the following formula:

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(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

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\Delta K=W_T         (3)

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W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

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d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

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