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kipiarov [429]
3 years ago
13

A man paddles a canoe at 6 km per hour. If he paddles on a river with a current of 6 km per hour, what is the speed of the canoe

if it heads: Upstream? Directly across the river?

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is  8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]

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A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of
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m = 2.23 \times 10^{-32} kg

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

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we know that

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Gravitational forceF= \frac{Gm m}{d^2}

we know that

v = \frac{2\pi R}{T}

solving for

R = \frac{vT}{2\pi}

F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}

F = G\times \frac{\pi m}{(vT)^2}

a = \frac{v^2}{\frac{vT}{2\pi}}

a = \frac{2\pi v}{T}

we know that

f =ma

G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}

solving for m

m = \frac{2Tv^3}{\pi G}

m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}

m = 2.23 \times 10^{-32} kg

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Hence, the answer is 3.

8 0
3 years ago
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