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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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The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

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Determine which the following are defined Whenever they occur denctes multivariable function and denotes vector field both of wh
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Yes, all of the above are defined whenever they occur and denote multivariable functions and vector fields which are twice continuously differentiable on a common domain.

What is vector field?
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The first expression, V . (F . Vf), denotes a multivariable function which is the dot product of a vector V and the composition of a function F with a vector field Vf.

The second expression, V . (V x Vf), denotes a multivariable function which is the dot product of a vector V and the cross product of a vector V and a vector field Vf.

The third expression, V x (V . F), denotes a vector field which is the cross product of a vector V and the dot product of a vector V and a function F.

The fourth expression, V x (V x f), denotes a vector field which is the cross product of a vector V and the cross product of a vector V and a function f.

The fifth expression, V x (V x F), denotes a vector field which is the cross product of a vector V and the cross product of a vector V and a function F.

The sixth expression, V . (V x F), denotes a multivariable function which is the dot product of a vector V and the cross product of a vector V and a function F.

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