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Sever21 [200]
3 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]3 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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Explanation:

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A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o
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Correct question is;

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to √2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s²)

Answer:

x(t) = 7te^(-2t√2)

Explanation:

We are given;

Weight; W = 8 lbs

mass; m = W/g

g = 32 ft/s²

Thus;

m = 8/32

m = ¼ slugs

From Newton's second law we can write the equation as;

m(d²x/dt²) = -kx - β(dx/dt)

Rearranging this, we have;

(d²x/dt²) + (β/m)(dx/dt) + (k/m)x = 0

Where;

β is damping constant = √2

k is spring constant = W/s

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Thus,we now have;

(d²x/dt²) + (√2/(¼))(dx/dt) + (2/(¼))x = 0

>> (d²x/dt²) + (4√2)dx/dt + 8x = 0

The auxiliary equation of this is;

m² + (4√2)m + 8 = 0

Using quadratic formula, we have;

m1 = m2 = -2√2

The general solution will be gotten from;

x_t = c1•e^(mt) + c2•t•e^(mt)

Plugging in the relevant values gives;

x_t = c1•e^(mt) + c2•t•e^(mt)

At initial condition of t = 0, x_t = 0 and thus; c1 = 0

Also at initial condition of t = 0, x'(0) = 7 and thus;

Since c1 = 0, then c2 = 7

Thus,equation of motion is;

x(t) = 7te^(-2t√2)

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