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Leno4ka [110]
3 years ago
5

A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds?​

Physics
1 answer:
QveST [7]3 years ago
3 0

Explanation:

F = ma

F = m Δv / Δt

F = (1500 kg) (0 m/s − 30 m/s) / 3 s

F = -15,000 N

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If the distance between two masses is tripled, the gravitational force between changes by a factor of
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A. 1/9

Explanation:

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F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

\frac{1}{r^2}=\frac{1}{3^2}=\frac{1}{9}

6 0
3 years ago
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Gasoline burns inside a car’s engine. how does this fuel enable a car to move?
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8 0
3 years ago
A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), Fig. 5–39.
Trava [24]

Answer:

1742.24106 revolutions per day

Explanation:

v = Velocity

d = Diameter = 1.1 km

r = Radius = \dfrac{d}{2}=\dfrac{1.1}{2}=0.55\ km

g = Acceleration due to gravity = 9.81 m/s²

g = 0.9 g

The centrifugal force will balance the gravitational force

F_c=mg\\\Rightarrow \dfrac{mv^2}{r}=m0.9g\\\Rightarrow v=\sqrt{\dfrac{0.9gmr}{m}}\\\Rightarrow v=\sqrt{0.9gr}\\\Rightarrow v=\sqrt{0.9\times 9.81\times 0.55\times 10^3}\\\Rightarrow v=69.68464\ m/s

\dfrac{1}{T}=\dfrac{v}{2\pi r}\\\Rightarrow \dfrac{1}{T}=\dfrac{69.68464}{2\pi 0.55\times 10^3}\times 24\times 60\times 60\\\Rightarrow \dfrac{1}{T}=1742.24106\ rev/day

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6 0
3 years ago
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Alla [95]
Answer:

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4 years ago
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