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2 years ago

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids?A. 17.2B. 0.0581C. 0.241D. 4.15

1 answer:

tekilochka [14]2 years ago

4 0

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

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b) Equal to 243 N.

Explanation:

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The action force is that of the pull by Harry acting on the car.
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2 years ago

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

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Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

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We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

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So,

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2 years ago

Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat

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Answer:

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1 year ago

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Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

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Now y = 2500 ft

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