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Leokris [45]
2 years ago
9

We know the moon circulates the Earth. Suppose the mass of the Earth and moon are 5.9742 x1024 kg and 7.36 x 1022 kg, whereas th

e radius of the orbit is 382171 km. What will the gravitational force between the Earth and the moon?
Physics
1 answer:
algol [13]2 years ago
4 0

Answer:

The gravitational force between two objects, one of mass M1 and the other of mass M2, is:

F = G*M1*M2/R^2

Where G is a constant:

G = 6.6x10^11 m^3*/(kg*s^2)

And R is the distance between the two objects.

M1 = 5.9742x10^24 kg

M2 = 7.36x10^22 kg

R = 382171 km = 382171000 m

Then the gravitational force is:

F = 6.6x10^-11 m^3*/(kg*s^2)*(5.9742x10^24 kg)*(7.36x10^22 kg)/(382171000 m)^2

F = 1.987x10^20 N

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Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad
mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = \rhoV

V = Volume =\dfrac{4}{3}\pi r^3

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

\rho = Density = 19 g/cm³

c = Speed of light = 3\times 10^8\ m/s

A = Area = \pi R^2

I = Intensity = \dfrac{P}{4\pi r^2}

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}

According to the question

F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W

The power required of the light source is 468449163762.0812 W

4 0
2 years ago
What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?
densk [106]
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get 
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
7 0
2 years ago
Read 2 more answers
Which of the following four circuit diagrams best represents the experiment described in this problem?
Valentin [98]

We don't see any circuit diagrams.  

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

6 0
2 years ago
7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr
wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polar
loris [4]

Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

For the second polarizer, the intensity is equal:

I_{2} =\frac{I_{o} }{2} (cos\theta )^{2} =\frac{I_{o} }{2} (cos90)^{2} =0

5 0
2 years ago
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