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tatuchka [14]
3 years ago
9

A tourist leaves Pittsburgh and travels 60 miles South on I-79 in 45 minutes. They exit after Wheeling and head west on I-70 and

drive an additional 100 miles in 1 hour and 15 minutes. What is the tourist’s average speed?
Physics
1 answer:
attashe74 [19]3 years ago
8 0

Answer:

<em>The tourist's average speed was 80 mph</em>

Explanation:

<u>Average Speed </u>

If an object travels a distance d in a time t regardless of the direction, the average speed is the quotient of the distance over the time:

\displaystyle v=\frac{d}{t}

The tourist leaves Pittsburgh and travels 60 miles South in 45 minutes. Then he heads West and travels 100 miles in 1 hour and 15 minutes.

The total distance traveled is:

d = 60 miles + 100 miles = 160 miles

The total time is:

t = 45 minutes + 1 hour + 15 minutes = 2 hours

The average speed is:

\displaystyle v=\frac{160}{2}

v = 80 mph

The tourist's average speed was 80 mph

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Answer:

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Explanation:

The length of the slide is

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So the height of the child when he is at the top of the slide is (with respect to the ground)

h = L sin \theta = (2.68 m)sin 25.0^{\circ}=1.13 m

The potential energy of the child at the top is given by:

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Substituting,

U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J

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h = 0

So the potential energy is also zero: U = 0 J.

This means that the change in potential energy as the child slides down is

\Delta U = 0 J - (1654.6 J) = -1654.6 J

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Distance,d=0.41 m

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Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

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F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

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