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Sever21 [200]
2 years ago
7

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

ry itself just below the surface.
Physics
1 answer:
MArishka [77]2 years ago
6 0

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

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A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks
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The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp
olganol [36]

Answer:

Ft

Explanation:

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Initial velocity=u=0

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v=0+\frac{F}{m}t=\frac{F}{m}t

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Using the formula

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6 0
2 years ago
A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi
KIM [24]

The stone reaches the wall at a height of <u>1.62 m</u>.

The stone lands at a point <u>24.5 m</u> from the point of projection.

The stone is projected horizontally with a velocity u at a height <em>h</em> from the ground. The wall is located at a distance <em>x</em> from the point of projection. The stone takes a time <em>t</em> to reach the wall and in the same time the stone falls a vertical distance <em>y</em>.

The horizontal distance <em>x</em> is traveled with a constant velocity <em>u</em>.

x=ut

Calculate the time taken <em>t</em>.

t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s

The stone's initial vertical velocity is zero. It falls through a distance <em>y</em> in the time <em>t</em> under the action of acceleration due to gravity <em>g</em>.

y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m

The height  <em>h₁ </em>of the stone above the ground when it reaches the wall  is given by,

h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m

When the stone reaches the wall, its height from the ground is <u>1.62m.</u>

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance <em>R</em> from the point of projection. If the time taken by the stone to reach the ground is <em>t₁, </em>then,

h=\frac{1}{2} gt_1^2

Calculate the time taken by the stone to reach the ground.

t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s

The horizontal distance traveled by the stone is given by,

R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0
2 years ago
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