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Helen [10]
3 years ago
10

he adventurous robot M.A.N.D.I. is orbiting Saturn’s moon Dione. She wants to cause an impact with themoon to kick up some of th

e surface dust so that she can do a spectral analysis of it. She tosses a steel ball bearing inthe opposite direction of her orbital velocity with just the right impulse to make the ball come to a dead stop. That’sthe back story. The actual problem starts here: The ball, starting with zero velocity, falls straight down to the surfaceof the moon. If the moon has a radius of5.61×103m and a mass of1.10×1021kg, and if the ball bearing starts at analtitude of2.73×103m above the surface of the moon, how fast will it be going when it hits the surface? Note thatthe gravitational constantG= 6.67408×10−11N m2/kg2.
Physics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

v = 2.928 10³ m / s

Explanation:

For this exercise we use Newton's second law where the force is the gravitational pull force

         F = ma

         a = F / m

Acceleration is

        a = dv / dt

        a = dv / dr dr / dt

        a = dv / dr v

        v dv = a dr

We substitute

       v dv = a dr

       ∫ v dv = 1 / m G m M ∫ 1 / r² dr

We integrate

       ½ v² = G M (-1 / r)

We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon

       v² = 2G M (- 1 / R +2.73 10³+ 1 / R)

         

We calculate

       v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61  - 10⁻³ /(5.61 + 2.73))

       v² = 14.6828 10⁷ (0.1783 -0.1199)

       v = √8.5748 10⁶

       v = 2.928 10³ m / s

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wariber [46]

Answer:

PE=81.755\, J

Explanation:

Given that:

  • mass of meteoroid, m=7.8\times 10^8 \,kg
  • radial distance from the center of the planet, R= 2.8\times 10^7 m
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<u>For gravitational potential energy we have:</u>

PE=G\frac{M.m}{R}

substituting the respective values:

PE=6.67\times 10^{-11}\times \frac{4.4\times 10^{25}\times 7.8\times 10^8}{2.8\times 10^7}

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5 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

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so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

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So we can use angular momentum conservation about the hinge point

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If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes d
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Moving company uses a machine to raise a 900 Newton refrigerator to the second floor of a building machine consists of a single
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3 years ago
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