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Romashka [77]
2 years ago
12

PLS HELP ASAP :(( You observe a car parked on the side of the road. If the car started to move, Which conclusion could you make?

Physics
2 answers:
Nostrana [21]2 years ago
7 0

Answer:

The first option

Explanation:

Nadya [2.5K]2 years ago
5 0

Answer:

A (the first one)

Explanation:

An object is stationary when both forces acting on it are balanced. As soon as the car started moving, the driver had placed extra force on the accelerator and therfore the forces became unbalanced and the car moved

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As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period
Hunter-Best [27]
Periodic time is the time taken for one complete oscillation by a body in circular motion. In this case the merry-go round takes 2 minutes to cover 15 complete oscillations. 2 Minutes = 120 seconds
Hence, 15 oscillations takes 120 secs
         thus 1 oscillation takes 120/15 = 8 seconds
therefore the period of the merry-go-round = 8 seconds
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How fast did Newton decide a cannon ball shot out in order to escape Earth’s gravity?
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Plugging in for the Earth's mass and for G, we have 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour

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2 years ago
How many amps of current are there in a system with 12 volts and 6 ohms?
Gemiola [76]
V = IR
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2 years ago
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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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