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Galina-37 [17]
3 years ago
10

A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire

ction that puts him directly across the river from the starting point, and then he returns in a direction that puts him back at the starting point in the shortest time possible. The travel time for the man is?
Physics
1 answer:
gladu [14]3 years ago
3 0

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

V_T^2=v_N^2-v_W^2

V_T = \sqrt{v_N^2-v_W^2}

Travel north 2mph and west to 1mph, then,

V_T = \sqrt{2^2-1^2}

V_T = \sqrt{3}

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

v= \frac{\Delta x}{t}

t = \frac{\Delta x}{v}

t = \frac{2*(1mile)}{\sqrt{3}mph}

t = 1.15hour

Therefore the total travel time for the man is 1.15hour.

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Answer:

Both experienced the same magnitude impulse

Explanation:

This is because, the impulse force is internal to the system of both the tennis ball and the bowling ball. It is an action-reaction pair. So, the force exerted on the tennis ball by the bowling ball equals in magnitude to the force exerted by the tennis ball on the bowling ball although, they are in opposite directions. This, both experienced the same magnitude impulse.

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planetary accretion is the term that is used

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A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.
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Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

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Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

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2 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
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Answer:

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Explanation:

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\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

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