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GrogVix [38]
3 years ago
13

You hold a 0.125 kg glider A and a 0.500 kg glider B at rest on an air track with a compressed spring of negligible mass between

them. When you release the gliders, the spring pushes them apart so that they move in opposite directions. When glider A has moved 0.960 m to the left from its starting position, how far to the right from its starting position has glider B moved?
Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

Magnitude |VB| = 0.32 m/s

and

Direction is to the left

Explanation:

given data

glider A mass = 0.125 kg

glider B mass = 0.500 kg

glider A move = 0.960 m to the left

solution

we use here Conservation of momentum  that is

M1 × U1 + M2 × U2 =M1 × V1 + M2 × V2    ........................1

put here value and we get

0.125 × 0 + 0.500 × 0 = 0.125 × 0.960 + 0.375 × VB

VB = −0.32  m/s

Magnitude |VB| = 0.32 m/s

and

Direction is to the left

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A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

8 0
3 years ago
An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh
melomori [17]

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.


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3 years ago
An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how
Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

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Using first equation of motion to find it as :

v=u+at

a is the acceleration, here, a=\mu g

0=u+\mu gt

t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

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