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?

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:

where

m is the mass

a is the acceleration

In this problem, we have:

- The net force can be written as

where

F = 40 N is the horizontal forward push

is the force of friction

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

By combining all the equations together, we find:

So, the force of friction is 40 N.

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explanation:

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :