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ANTONII [103]
3 years ago
15

5. How much force is required to accelerate a 12kg mass at 5m/s??

Chemistry
2 answers:
alisha [4.7K]3 years ago
5 0

Answer:

Therefore the force required to accelerate 12 kg mass at 5 m/s²  is 60N

Explanation:

<u><em>How much force is required to accelerate a 12kg mass at 5m/s?</em></u>

Using newton's law of motion;

F = mass × acceleration

mass is given to be 12 kilogram and acceleration is given to be 5 meter per square second.

So we will substitute the value into our equation;

F = mass × acceleration

   =12 kg  ×  5 m/s²

   = 60 kg m/s

F =  60 kg m/s²

But kg m/s²= N

The unit for measuring force is N

F =  60 N

Therefore the force required to accelerate 12 kg mass at 5 m/s²  is 60N

gregori [183]3 years ago
3 0

Answer:

Force = Mass x Acceleration

12 x 5 = 60

60 Newton’s is your final answer

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Calculate the volume of 0.684mol of carbon dioxide at s.t.p. show working
solong [7]

Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.684

R = gas constant = 0.0821Latm/Kmol

T =temperature =273K   (at STP)

V=\frac{nRT}{P}

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6 0
3 years ago
A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the vo
kozerog [31]

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

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0.346*777 = (463*V2) / (777)*777

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268.842/463 = (463*V2)/463

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Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

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