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yKpoI14uk [10]
3 years ago
5

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha

d a mass of 130kg, how much would you weigh when standing on the surface of this asteroid?B. If you could walk on the surface of this asteroid, what minimum speed would you need to launch yourself into an orbit just above the surface of the asteroid?
Physics
1 answer:
WITCHER [35]3 years ago
3 0

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

F=\frac{GMm}{R^2}

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

\frac{GMm}{R^2}=\frac{mv^2}{R}

where

v is the speed of the astronaut

Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)}{2000 m}}=2.41 m/s

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Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

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mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

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⇒  Ycm = (ms*D + mp*0) / (ms + mp)

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3 years ago
A car travels 82 meters do North and 14 seconds the car turns around and travels 44 m due south in four seconds what is the magn
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Answer:

2.11 m/s

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The velocity is positive, so it is 2.11 m/s north.  The magnitude of the velocity is 2.11 m/s.

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Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
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Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

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D ) Given t = 10 years , t₁ = ? v = .4c

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t₁= 10.9  years

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A block slides down an incline plane that makes a 30 degree angle with the
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Hi there!

We know that:

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Let the positive direction be directed in the direction of the block's acceleration, which is downward.

Thus:

ΣF = Mgsinθ - μMgcosθ

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a = gsinθ - μgcosθ

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a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²

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