Potential energy is energy due to an object's height above the ground.
Potential energy = mass x gravity x height
Kinetic energy is energy due to the motion of the object.
Kinetic energy = 1/2 x mass x velocity²
1.
The ball is not moving and is at a height above the ground so it has only potential energy.
P.E = 2 x 9.81 x 40
P.E = 784.8 J
2.
The ball is moving and has a height above the Earth's surface so it has both kinetic and potential energy.
P.E = same as part 1 = 784.8 J
K.E = 1/2 x 2 x 5²
K.E = 25 J
3.
The ball has no height above the Earth's surface and is moving so it has only kinetic energy.
K.E = 1/2 x 2 x 10²
K.E = 100 J
4.
50000 = 1/2 x 1000 x v²
v = 10 m/s
5.
39200 = 200 x 9.81 x h
h = 20.0 m
6.
12.5 = 1/2 x 1 x v²
v = 5 m/s
98 = 1 x 9.81 x h
h = 10.0 m
<span>Temperature is defined as the rate at which molecules move or vibrate
</span>
621.4L
Explanation:
Given parameters:
Initial volume = 547L
Initial temperature = 331K
Final temperature = 376K
Unknown:
Final volume = ?
Solution:
The appropriate gas law to use is the Charles's law.
The Charles's law shows the relationship between the volume and temperature of a gas under constant pressure.
The law states that "The volume of a fixed of a gas varies directly as its absolute temperature if the pressure is constant".
Mathematically;
V₁ is the initial volume
T₁ is the initial temperature
V₂ is the final volume
T₂ is the final temperature
Since the unknown is the final volume, we make it the subject of the expression;
V₂ = 
V₂ = 621.4L
learn more:
Boyle's law brainly.com/question/8928288
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Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
