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kow [346]
2 years ago
9

Suppose the moon were held in its orbit not by gravity but by a massless cable attached to the center of the earth.Part AWhat wo

uld be the tension in the cable? Use the table of astronomical data inside the back cover of the textbook.Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
MAXImum [283]2 years ago
3 0

Answer:

1.98\cdot 10^{20}N

Explanation:

The tension in the cable would be exactly equal to the force of gravity between Moon and Earth, which is given by:

F=G \frac{mM}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m=7.35\cdot 10^{22} kg is the mass of the Moon

M=5.97\cdot 10^{24} kg is the mass of the Earth

r=3.84\cdot 10^8 m is the distance between Moon and Earth

Substituting numbers into the equation, we find

F=(6.67\cdot 10^{-11}) \frac{(7.35\cdot 10^{22})(5.97\cdot 10^{24})}{(3.84\cdot 10^8)^2}=1.98\cdot 10^{20}N

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Georgia [21]

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

5 0
2 years ago
The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect
Diano4ka-milaya [45]
Rx= 3.5 km

Ry= 2.9 km
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2 years ago
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A negatively charged glass rod is brought near a neutral table tennis ball. What will happen to the neutral table tennis ball?.
Zina [86]

The neutral table tennis ball will become polarized, with positive charges toward the glass rod. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

7 0
2 years ago
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a cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.- kilogram cannon, initially at rest, on a horizontal,
nirvana33 [79]

(a) According to Newton’s third law, for every action there is an equal yet opposite reaction. In this case, the cannonball is affected by 8.0x10^3 newtons in one direction. Now the cannon must also be affected by the same amount of net force on the opposite direction.

- 8.0x10^3 Newtons (the negative symbol only shows that it is opposite direction)

 

(b) We solve this using the formula:

F = m a

8.0x10^3 N = (1.0 kg) a

a = 8,000 m/s^2

8 0
2 years ago
Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

5 0
2 years ago
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