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sweet [91]
3 years ago
10

A 4kg block and a 2kg block can move on horizontal frictionless surface. The blocks are accelerated by a +12-N force that pushes

the large block against the smaller one.'
a) Determine the force that the 2kg block exerts on the 4kg block

b) Determine the force that the 4kg block exerts on the 2kg block.

Physics
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

a) -4 N

b) +4 N

Explanation:

Draw a free body diagram for each block.

For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.

For the small block, there is 1 force, F pushing to the right.

There are also weight and normal forces in the vertical direction, but we can ignore those.

Sum of forces on the large block in the x direction:

∑F = ma

12 − F = 4a

Sum of forces on the small block in the x direction:

∑F = ma

F = 2a

2F = 4a

Substitute:

12 − F = 2F

12 = 3F

F = 4

The small block pushes on the large block 4 N to the left (-4 N).

The large block pushes on the small block 4 N to the right (+4 N).

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temperature on left side is 1.48 times the temperature on right

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We know that

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n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

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6 0
3 years ago
An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.
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Answer:

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Explanation:

<u>Given the following data;</u>

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a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

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ira [324]

Answer:

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2 years ago
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