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r-ruslan [8.4K]
3 years ago
12

The moon's mass is 7.34x10-kg and it is 3.8x10m away from earth. Calculate the gravitational force of attraction between earth a

nd moon
Physics
2 answers:
Nana76 [90]3 years ago
8 0
Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.

<span>The formula for force of attraction between any two bodies in the universe
F  =  GMm / r^2.      (Newton's Universal law of Gravitation).

G = Universal gravitational constant, G = 6.67 * 10 ^ -11  Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22  kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
   = 3.8 x 10^8 m.

(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).


So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
  Punch it all up in your calculator.
 
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.

F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!</span>




Ivahew [28]3 years ago
8 0
Using the gravity equation: force of gravity = (G*mass of object 1*mass of object 2)/distance between the objects^2, we can plug in the masses of the objects (5.97x10^24kg for earth, and 7.34x10^22kg for the moon) and the distance between the objects (3.8x10^8metres) and (6.67x10^-11 gravitational constant) for G, we get (6.67x10^-11*5.97x10^24*7.34^22)/(3.8x10^8)^2 which equals =2.02x10^20 which is the force of gravity
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Ou are given a 25.3 µf capacitor that is connected to a 13.0 v dc power supply. what will be the charge that is stored on this c
Black_prince [1.1K]

charge stored in the capacitor=3.29 x 10⁻⁴ C

Explanation:

we use the formula

Q= C V

Q= charge

C= capacitor=25.3 μF= 25.3 x 10⁻⁶ F

V= voltage= 13 V

Q=(25.3 x 10⁻⁶ ) (13)

Q= 3.29 x 10⁻⁴ C

5 0
3 years ago
When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d
Mice21 [21]

The gravitational acceleration at any distance r is given by

g=  \frac{GM}{r^2}

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is r_e=6.37 \cdot 10^6 m, so the meteoroid is located at a distance of:

r=r_e+2.60 r_e =3.60 r_e =  2.29 \cdot 10^7 m

And by substituting this value into the previous formula, we can find the value of g at that altitude:

g=  \frac{GM}{r^2} =  \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2

5 0
3 years ago
Read 2 more answers
2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste
grin007 [14]

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

I = m L^2        

where L is the length of road, we will take whole length of rod because mass is at  the end of it.      

I = 1.5 \times 0.85^2  

I = 1.084 kg.m²                        

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

8 0
2 years ago
If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine
pickupchik [31]

Answer:

τ = 132.773 lb/in² = 132.773 psi

Explanation:

b = 12 in

F = 60 lb

D = 3.90 in (outer diameter)  ⇒ R = D/2 = 3.90 in/2 = 1.95 in

d = 3.65 in (inner diameter)  ⇒ r = d/2 = 3.65 in/2 = 1.825 in

We can see the pic shown in order to understand the question.

Then we get

Mt = b*F*Sin 30°

⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in

Now we find ωt as follows

ωt = π*(R⁴ - r⁴)/(2R)

⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)

⇒ ωt = 2.7114 in³

then the principal stresses in the pipe at point A is

τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)

⇒ τ = 132.773 lb/in² = 132.773 psi

7 0
3 years ago
HELP ASAP!!
andriy [413]

Answer:

This is false becuase different object weigh different

Thank you!

Explanation:

3 0
2 years ago
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