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Gelneren [198K]
2 years ago
5

An object of mass 10 kg is moving along positive x- axis with velocity 5 m/s. At some point it breaks into two parts of 6 kg and

4 kg. 6 kg moves making 600 with x –axis and 4 kg moves making -300 with axis. What is velocity of their centre of mass after breaking?
Physics
1 answer:
vichka [17]2 years ago
6 0

Explanation:

Correct option is

B

−3m/s

2

Given,

m=10kg

u=10m/s

v=−2m/s

t=4sec

The acceleration is rate of change of velocity,

a=

t

v−u

a=

4

−2−10

a=

4

−12

=−3m/s

2

The correct option is B.

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A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th
Liono4ka [1.6K]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, m=\dfrac{F}{a}

m=\dfrac{70\ N}{7.5\ m/s^2}

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

W=m\times g

W=9.34\ kg\times 1.62\ m/s^2

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or

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So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

3 0
4 years ago
A firefighter with a weight of 756 N slides down a vertical pole with an acceleration of 2.96 m/s2, directed downward. What are
Katen [24]

Answer:

a) F = 527.65 N, Force applied is upwards.

b)F = - 527.65 N, where, negative sign depicts Force is applied downwards.

Explanation:

Data provided:

Weight of the firefighter = 756 N

Mass of the firefighter = 756/9.8 = 77.14 Kg

Acceleration, a = 2.96 m/s²

a) In the absence of the pole the firefighter would have been moving down with an acceleration of 9.8 m/s (i.e the acceleration due to the gravity), but due to the presence of the pole the acceleration of the firefighter has been reduced. thus, a force is applied by the pole on the firefighter to reduce the acceleration.

therefore, we have

F = ma(net) = 77.14 × (9.8-2.96) = 527.65 N, Force applied is upwards.

B) According to the Newton's third law, the force will be equal and opposite to the force in the part a)

thus, we have

F = - 527.65 N

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