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Zolol [24]
3 years ago
8

The double inclined plane supports two blocks A and B, each having a weight of 10 lb. If the coefficient of kinetic friction bet

ween the blocks and the plane is 0.1, determine the acceleration of each block. (The blocks are connected to each other by a string and pulley system so their accelerations are equal, block on the left is on an incline of 60 degrees and the block on the right is on an incline of 30 degrees.)
Physics
1 answer:
Vladimir79 [104]3 years ago
8 0

Answer:

Explanation:

Left block is on surface with higher inclination so it will go down . If T be tension

For motion of block A ,

net force = mgsin60 - (T + mg cos 60 x μ ) , μ is coefficient of friction .

ma =  mgsin60 - T -  mg cos 60 x .1

10a = 277.13  - T - 16

= 261.13 - T

T = 261.13 - 10a

For motion of block B

T - mg sin30 - mgcos30 x μ = ma

T- 160 - 27.71 = 10 a

261.13 - 10a - 160 - 27.71 = 10a

73.42 = 20a

a = 3.67 ft / s²

common acceleration = 3.67 ft / s²

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Answer:

a)V= 0.0827 m³

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Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

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81.5 = V x 985

V= 0.0827 m³

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Area ,A= 4.5 x 10⁻² m²

The Pressure P

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t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

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Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

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t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

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