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Hitman42 [59]
3 years ago
15

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

n a 38.0-kg object placed midway between them.
Physics
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

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The leader Fulgencio Batista maintained control of Cuba from 1934 to 1959.

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3 years ago
on a distant planet a freely falling object has an acceleration of 20 m/s.^2 what speed will a body dropped from rest on his pla
N76 [4]

Answer:

30 m/s

Explanation:

Vf = ?

Vi = 0 m/s

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Plug those values into the equation: Vf = Vi + at

Vf = 0 + (20)(1.5)

Vf = 30 m/s

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Which is a concern about mining for uranium? Heated water could be released into the environment. Dust released in the air could
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Answer: Dust released in the air could be radioactive.

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Thus, the correct answer is dust released in the air could be radioactive is a concern about mining Uranium.

3 0
3 years ago
Read 2 more answers
A force of 70 N is applied to a 28 kg rock causing it to slow down from 25 m/s to 15 m/s, a change in velocity of 10 m/s. How lo
maksim [4K]

Answer: 4 s

Explanation:

Given

The applied force is 70 N

mass of the rock is 28 kg

initial velocity u=25\ m/s

final velocity v=15\ m/s

Deceleration provided by force is

a=-\dfrac{70}{28}=-2.5\ m/s^2

using the equation of motion

v=u+at\\\Rightarrow 15=25-2.5t\\\Rightarrow 2.5t=10\\\Rightarrow t=4\ s

7 0
3 years ago
An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular electric and a magnetic fields a
Cloud [144]

Answer:

magnetic field will allow the electron to go through 2 x 10^{4}  T k

Explanation:

Given data in question

velocity = 5.0 × 10^{7}

electric filed = 10^{4}

To find out

what magnetic field will allow the electron to go through, undeflected

solution

we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions

so we can also say

F(net) = Fe + Fb i.e. = 0

q V B +  q E = 0

q will be cancel out

10^{4}j + 5e + 7i × B =  0

B = 2 x 10^{4}  T k

3 0
3 years ago
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