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erik [133]
3 years ago
8

What is the unit of G in the F=Gm1m2/r^2​

Physics
2 answers:
Sloan [31]3 years ago
7 0
Where: F is the force between the masses; G is the gravitational constant (6.674×10−11 N. (m/kg)2);
kobusy [5.1K]3 years ago
6 0


G
has the SI units
m
3
k
g
⋅
s
2
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How much carbon dioxide (co2) does natural gas emit compared with other fossil fuel
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4 years ago
A 30kg boxed is pushed with a force of 20N. What is the boxes acceleration. Please show work
kozerog [31]

Answer:

<h3>The answer is 0.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{20}{30}  =  \frac{2}{3}  \\  = 0.6666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

8 0
3 years ago
A planet orbits a sun in a clockwise elliptical orbit as shown in the diagram below
STatiana [176]

Answer:

Greatest gravitational energy is at "C".

The planet has to do work "against" the field to get to "C".

Also, if m v R (angular momentum) is constant then as R increases v must decrease for this term to be constant and KE = 1/2 M v^2  must decrease also to get to point C.

3 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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