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Naddik [55]
3 years ago
10

You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club an

d a golf ball.
Part A

(How fast would you need to hit the ball horizontally so it goes into a circular orbit?)

Express your answer with the appropriate units.

Part B

(If you hit the ball vertically with the same speed, to what height above you would it rise?)

Express your answer with the appropriate units.
Physics
1 answer:
Anna71 [15]3 years ago
6 0

Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

acceleration is centripetal

        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          Em_{f} = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m

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guapka [62]

Answer:

Part a)

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

mv + MV = 0

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR

now we have

\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR

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V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

and another block has speed

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

7 0
3 years ago
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacit
Roman55 [17]

Answer:

a) Energy stored in the capacitor, E = 1.0125 *10^{-3} J

b) Q = 45 µC

c) C' = 1.5 μF

d)  E = 6.75 *10^{-4} J

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J

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Sherry and Ellen are talking about the factors that affect light scattering in the atmosphere. Sherry says one factor is the len
Delvig [45]

Sherry who says one factor is the length of the path of sunlight is correct.

<h3>Factors affecting light scattering</h3>

There are two main factors which affects light scattering, and they include the following;

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length of the path of sunlight is equivalent to wavelength of the light.

Thus, we can conclude that Sherry who says one factor is the length of the path of sunlight is correct.

Learn more about light scattering here: brainly.com/question/1381101

#SPJ1

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Answer:

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