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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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Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

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In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

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B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

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6 0
3 years ago
A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the
shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

Force, F = ma

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

a = 0.47 * 9.8 m/s^2

a = 4.606m/s^2

We know,

v^2 - u^2 = 2as

where, v is the final velocity

           u is the initial velocity

           a is the acceleration

           s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s

The car travels at a speed of 25m/s.

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An object will not change its motion unless an outside force acts on it true or false
Gre4nikov [31]

Answer:

True!

Explanation:

Newton's First Law :)

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