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sp2606 [1]
3 years ago
12

What is the equation for the law of universal gravitation?

Physics
2 answers:
Andrei [34K]3 years ago
8 0
Newton's law of Universal Gravitation says that one particle attracts the rest of every other particle in the universe using a force that is directly equivalent to the product of their masses and inversely equivalent to the square of the distance between them.
mars1129 [50]3 years ago
7 0
Equation for law of Universal Gravitation is:

F = G. m₁ m₂ / r²

Where, m₁, m₂ = masses of the object
r = distance between them
G = Gravitational Constant (6.67 * 10⁻¹¹ m³ Kg⁻¹ s⁻²

Hope this helps!
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A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential h
Sloan [31]

Answer:

77.88 lbm/ft³

Explanation:

Given,

Specific gravity, SG = 1.25

Density of water, ρ = 62.30 lbm/ft³

density of the fluid =

   = S.G x ρ_{water}

   = 62.30 x 1.25

   = 77.88 lbm/ft³

Density of the fluid is equal to 77.88 lbm/ft³

3 0
3 years ago
The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2
dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

P=\dfrac{V^2}{R}

Substitute all the values,

P=\dfrac{3^2}{2}\\\\P=4.5\ W

So, the power of this circuit is equal to 4.5 watts.

5 0
3 years ago
Jupiter is made of gas(like Saturn, Uranus and Neptune). What would happen to the strength of gravity if you
garik1379 [7]

Answer:

a) The strength of gravity decreases if one moved away from Jupiter

b) The strength of gravity increases if one fell into Jupiter

Explanation:

The gravitational attraction is given by Newton law of gravitation as follows;

Force \ (strength) \ of \ gravity = \dfrac{G \times M \times m}{R^2}

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

M = The mass of Jupiter

m = The mass of the nearby body

R = The distance between the centers of Jupiter and the body

From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies

Therefore, as one moves away, R increases, and the strength of gravity reduces

Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.

7 0
3 years ago
A 900 N student runs up the stairs 3.5 m high in 12<br> seconds. How much POWER do they generate?
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Answer:

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4 0
3 years ago
The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\
lbvjy [14]
 it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt 
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x} 
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y} 
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ] 
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt 
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases 
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume 
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ] 
5 0
3 years ago
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