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2 years ago

6

How is motion affected by change in mass of an object and forces applied?

1 answer:

densk [106]2 years ago

6 0

Newtons second law says that the acceleration of an object (produced by a net force) is directly proportional to that magnitude of the net force. E.g. F = ma
where F is the net force of an object, m is mass and a is acceleration.
For example, if an object had a large mass, there would have to be more force in order to move it than if it was lighter.
In a linear motion, if you pushed two objects, one slightly larger than the other, with the same force, the acceleration of the smaller object would be bigger than the larger one. So the motion (change in position over time), of the larger object would be seen as lesser than the smaller one (in a situation where both forces are equal).

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2 years ago

A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

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mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

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3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

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$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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