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Mazyrski [523]
3 years ago
12

Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo

tion to calculate the amount of acceleration the object will experience. Then explain how the amount of acceleration will change if the net force or the mass of the object increases.
Chemistry
1 answer:
Luda [366]3 years ago
8 0

<u>Answer:</u> The acceleration of the object is 2m/s^2. If net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

<u>Explanation:</u>

Force is defined as the product of object's mass and acceleration.

Mathematically,

F=ma       ......(1)

or,

a=\frac{F}{m}      .....(2)

where,

F = Force exerted on an object = 60N

m = mass of an object = 30kg

a = acceleration of the object = ?

Putting values in above equation, we get:

a=\frac{60kg.m/s^2}{30kg}=2m/s^2

The acceleration of the car is 2m/s^2.

From equation 2, it is visible that acceleration is directly proportional to force. This means that \if force increases, acceleration also increases.

And acceleration is inversely proportional to mass of the object. This means that if mass increases, the acceleration decreases.

Hence, if net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

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Answer:

p_{H_2O}=2.00atm

Explanation:

Hello!

In this case, according to the following chemical reaction:

2H_2+O_2\rightarrow 2H_2O

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm

Best regards!

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