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tresset_1 [31]
2 years ago
14

How far from the earth must a body be along a line toward the sun so that the sun’s

Physics
1 answer:
Ivenika [448]2 years ago
4 0

Answer:

r = 1.63×10^5 mi

Explanation:

Let r = distance of object from earth

Rs = distance between earth and sun

Ms = mass of the sun

= 3.24×10^5 Me (Me = mass of earth)

At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,

Fs = Fe

GmMs/(Rs - r)^2 = GmMe/r^2

This simplifies to

Ms/(Rs - r)^2 = Me/r^2

(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2

Taking the reciprocal and then its square root, this simplifies further to

Rs - r = (569.2)r ----> Rs = 570.2r

or

r = Rs/570.2 = (9.3×10^7 mi)/570.2

= 1.63×10^5 mi

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Answer:

A.

Explanation:

NEAR THE CENTER OF TECTONIC PLATES.

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Consider an alien on a planet with an acceleration of gravity equal to 20 m/sec^2. If the alien's mass is 10 kg, how much does t
mr_godi [17]

Answer:

the weight of alien is 200 newtons

Explanation:

The computation of the alien weight is shown below:

Given that

Acceleration = 20m/sec^2

And, the mass is 10 kg

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Two charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges.For
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3 years ago
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l
deff fn [24]

Answer: 62 μT

Explanation:

Given

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Velocity of rod, v = 3.19 m/s

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Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

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8 0
2 years ago
Read 2 more answers
A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
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To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
2 years ago
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