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Levart [38]
3 years ago
12

How does The centripetal fotce change if the car has less Mass?​

Physics
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer:

The centripetal force acting on the car is proportional to the mass of the car.

Explanation:

Let,

The mass of the car be 'm'

The velocity of the car moving in the curved path be 'v'

The radius of the curved path be 'r'

According to physics, a body moving ion circular path experience a force directed along the radius of the path. This force is called centripetal force.

The formula for centripetal force is,

                                   <em>F = mv²/r</em>

Where,

                                      a = v²/r

So, if the mass of the car changes, the centripetal force also changes proportionally according to the above equation.

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Answer:

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When a wave strikes a new medium, three things can happen: reflection, transmission, and absorption. Consider an echo. Do you th
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Answer:

YES

Explanation:

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Although the direction of a sound changes but the echo sounds in the same way as the original sound.

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silicon is the principal component of glass and has properties of both metals and non-metals. this makes silicon A) Metal B) Non
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5 0
3 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N.
Likurg_2 [28]
The answers to the problem are as follows:

  <span>a) use f=ma. Mass of student is 836/9.8=85.3kg, force is 935-836=99N. Put it into the formula you will get a=99/85.3 

b)Use same formula. Force is 782-836=-54 so put it into formula a=-54/85.3 

c)Stopping would take longer as the acceleration is smaller 


I hope my answer has come to your help. God bless and have a nice day ahead!</span>
5 0
3 years ago
Read 2 more answers
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