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Whitepunk [10]
2 years ago
11

You are on the south bank of the river in your canoe you need to reach the north bank you know that you can row your canoe at 2

m/s in still water but the river is flowing 1 m/s due west, at what angle would respect to the north bank should you row your canoe at
Physics
1 answer:
Anna71 [15]2 years ago
3 0

Answer:

(θ) = 60°

Explanation:

Given:

Speed of canoe Vc = 2 m/s

Speed of River Vr = 1 m/s

Computation:

Vc (Cosθ) = Vr

2 (Cosθ) = 1

(Cosθ) =  1 / 2

(Cosθ) = (Cos60)

(θ) = 60°

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What is the frequency of a wave that has a period of 50 seconds?
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7 0
2 years ago
The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?
katovenus [111]
<h3><u>Question: </u></h3>

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

<h3><u>Answer:</u></h3>

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

<h3><u> Explanation: </u></h3>

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence F_{G} = F_{C}.

Gravitational force between Earth and Satellite: F_{G} = \frac{G \times M_e \times M_s}{R^2}

Centripetal force of Satellite :F_C = \frac{M_s \times V^2}{R}

Where G = Gravitational Constant

M_e= Mass of Earth

M_s= Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating  F_G = F_C, we get  

Speed of Satellite V =\frac{\sqrt{G \times M_e}}{R}

Thus the speed of satellite depends only on the mass of Earth.

6 0
3 years ago
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
2 years ago
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