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PolarNik [594]
3 years ago
14

A 10.0 kg crate is pushed with a horizontal force of 40 N. The crate moves at a constant velocity of 3.0 m/s. What is the value

of the friction force on the crate?
Chemistry
1 answer:
irga5000 [103]3 years ago
5 0

Answer:

40 N

Explanation:

Newton's second law of motion states that the net force acting on the crate is equal to the product between its mass and its acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have:

- The net force can be written as

\sum F = F - F_f

where

F = 40 N is the horizontal forward push

F_f is the force of friction

Also, we know that the crate moves at constant velocity, so its acceleration is zero:

a=0

By combining all the equations together, we find:

F-F_f=0\\F_f = F = 40 N

So, the force of friction is 40 N.

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
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Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

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At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

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[NO] at the equilibrium is 0.062 M

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For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

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