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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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Step2247 [10]

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

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here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

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\Delta p = \rho gh

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\Delta p = 1000*5 *9.81 = 49,050 kPa

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8 0
3 years ago
Read 2 more answers
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
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Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

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F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

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The amount of diffraction depends on the size of the obstacle and the wavelength of the wave.
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Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
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Answer:

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A = 2\times \frac{4}{\sqrt{3}}r

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Answer:

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