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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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you work at a Dave's donut shop the shop sells a dozen donuts for 6.99 dollars Dave redesigned the Box used to hold one dozen do
Fofino [41]

Answer:

A box of donuts would cost: b = 3.84 + 0.18f

First, we have to find the total cost of the donuts.

12 x 0.32 = 3.84

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The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity
n200080 [17]
<h2>Answer:g=9.79ms^{-2},A object of mass m at the surface of earth experiences a force mg</h2>

Explanation:

Let M be the mass of earth.

Let R be the radius of earth.

Let G be the universal gravitational constant.

Given,

M=5.96\times 10^{24}Kg

R=6.37\times 10^{6}m

G=6.67259 \times 10^{-11}Nm^{2}Kg^{-2}

Let g be the acceleration due to gravity.

Then,g=\dfrac{GM}{R^{2}}

g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}

g=9.79ms^{-2}

A object of mass m at the surface of earth experiences a force mg

3 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A²  ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J  = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

3 0
3 years ago
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