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dalvyx [7]
3 years ago
10

A) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a

distance of 3 km from its surface
b) if an astronaut popped out of a worm hole (at rest) at 3 km from the asteroid
i) how long would it take him to fall to the asteroid’s surface?
ii) how fast would he be traveling when he hit it? (Assume acceleration stays constant)
iii) if a year is 3.16 * 10^7 s, how many years would it take for the astronaut to reach the asteroid?
Please show work, thanks
Physics
1 answer:
MrMuchimi3 years ago
8 0

a) 1.96\cdot 10^{-16} m/s^2

The gravitational field strength near the surface of the asteroid is given by:

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the asteroid

R the radius of the asteroid

h is the distance from the surface

Substituting the data of the asteroid:

M=3.2\cdot 10^3 kg is the mass

R=30 km = 30000 m is the radius of the asteroid

h=3 km = 3000 m is the distance from the surface

We find

g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2

b) i)  5.53\cdot 10^9 s

The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

g=1.96\cdot 10^{-16} m/s^2

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

s=ut+\frac{1}{2}gt^2

where

s = 3 km = 3000 m is his displacement to reach the surface

u = 0 is his initial velocity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s

b) ii) 1.08\cdot 10^{-6} m/s

Again, we can use another suvat equation:

v=u+gt

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

Since we have

u = 0

t=5.53\cdot 10^9 s

g=1.96\cdot 10^{-16} m/s^2

The velocity of the astronaut at the surface will be

v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s

b) iii) 175 years

The duration of one year here is

T=3.16\cdot 10^7 s

And the time it takes for the astronaut to reach the surface of the asteroid is

t=5.53\cdot 10^9 s

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175

So, the astronaut will take 175 years to reach the surface.

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Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

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N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

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