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Vladimir [108]
3 years ago
14

10. If the gravitational force between two objects of equal mass is found to be 4.60*10^-6 N when the objects are 15 meters apar

t, what is the mass of each object?
Physics
1 answer:
liraira [26]3 years ago
5 0

Answer:

Explanation:

F = G*m1 * m2 / r^2

G = 6.67408 × 10-11 m3 kg-1 s-2

r = 15 m

F = 4.60 * 10^-6 N

m1 = m2 = m

F = 6.67 * 10^-11 * m^2 / 15^2

4.60 * 10^-6 = 6.67 * 10^-11 * m^2 / 225            Multiply both sides by 225

4.60*10^-6 * 225 = 6.67*10^-11 * m^2

.001035 =  6.67408 × 10-11 * m^2                     Divide by  6.67408 × 10-11

0.001035/ 6.67408 × 10-11 = m^2

m^2 = 15507755                                                Take the square root of both sides

m = sqrt(15507755   )

m = 3938 kg

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Vera_Pavlovna [14]

Answer:

1097.8 V/m

Explanation:

The equation that relates the intensity of an electromagnetic wave with the magnitude of the electric field is:

I=\frac{1}{2}c\epsilon_0 E^2

where

c is the speed of light

\epsilon_0 is the vacuum permittivity

E is the peak magnitude of the electric field

In this problem, we know the intensity:

I = 1600 W/m^2

So we can rearrange the formula to find E:

E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1600 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=1097.8 V/m

4 0
3 years ago
A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. T
nika2105 [10]

Answer:

0.955286  j

Explanation:

A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. Then a small explosion sends the module backward with speed 100.0 m/s relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft, Ki, increase because of the explosion?

M=500 kg,  m=440 kg

V=1000 m/s, v = 100 m/s

Let relative speed =Vs

Momentum rule says

(M+m)V=mVs+M(Vs-v)

 940(1000)=500(Vs-100)+440Vs

 940000=500Vs-50000+440Vs

 940Vs=940000+50000

 940Vs=990000

 Vs= 990000/940=1053.19 m/s

So, the module speed = Vs-v=1053.19-100=953.19 m/s

Fractional increase in KE is given by;

Total KE after explosion / He before explosion

=500(953.19)2+ 400(1053.19)2/ 940(1000)2= 0.955286

5 0
4 years ago
A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr
Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

brainly.com/question/15365772

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7 0
2 years ago
A square sheet of rubber has sided that are 20 cm long. If you stretch one side so that it is twice as long. What is the new are
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Answer:ignore this it was wrong

Explanation:

3 0
4 years ago
¿Tres ejemplos de rozamiento cinético?
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Answer:

Un automóvil en movimiento en una carretera.

Una piedra rodando por una colina.

Un hierro que se empuja a través del materia

4 0
3 years ago
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