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tino4ka555 [31]
2 years ago
12

If a 16 N net force makes an object accelerate at 4 m/s2, what is the mass of the object?

Physics
1 answer:
12345 [234]2 years ago
4 0

Answer:

4 kg

Explanation:

Acceleration = 4 m/s^2

Force = 16 N

Force = mass * acceleration

mass = force / acceleration

mass = 16 / 4

mass = 4 kg

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In physical changes no new materials are formed and the particles do not change apart from gaining or losing energy. ... Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas. Only their arrangement, energy and movement changes.

Explanation:

Hope this helps

7 0
3 years ago
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How long will it take a car to travel a distance of 1km if it has an average speed of 60kph​
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1÷60 h

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3 years ago
What force is required to accelerate a body with a mass of 15kilograms at a rate of
Paladinen [302]

The force required is

                     (15 kg) x (the acceleration, in m/s²)          newtons.

4 0
3 years ago
What change occurs to the mass of an object when an unbalanced force is applied to it
strojnjashka [21]
The mass of an object always stays the same since it is really just the amount of matter in an object so no matter the force applied, as long as the object does not lose or gain matter, the object stays the same
5 0
2 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
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