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3 years ago

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Set up the ion formation equations, with ionization energy values for each electron in the valence layer, of the atoms of the ch

emical elements below: a) Na z=11 b) Ca z= 20 c) Sr z 38 d) Li z= 3 e) Cs z= 55 f) Be z= 43

1 answer:

devlian [24]3 years ago

6 0

Explanation:

here is your answer.........

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If you use a horizontal force of 32.0 N to slide a 12.5 kg wooden crate across a floor at a constant velocity, what is the coeff

sergejj [24]

Answer:

Value of coefficient of kinetic friction is 0.26 .

Explanation:

Given:

Mass of wooden crate, m = 12.5 kg.

Horizontal force to keep the block moving with constant velocity, F = 32.0 N.

Since, the block is moving with constant velocity.

So, net force experience by it is zero.

Therefore, fore of friction is equal to applied force.

Now, force of friction , $F=\mu_kN$ ( here $\mu_k$ is coefficient of kinetic friction and N is normal force)

Therefore, $\mu_kN=\mu_k\times mg=\mu_k\times 12.5 \times 9.8=122.5\times \mu_k$

Now, both forces are equal.

$122.5\times \mu_k=32\\\mu_k=\dfrac{32}{122.5}=0.26$

The value of coefficient of kinetic friction is 0.26 .

Hence, it is the required solution.

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DedPeter [7]

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