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nikdorinn [45]
3 years ago
8

Place the notebook on a table, and push it a few inches with your finger. When it moves, it has kinetic energy. Which kind of fo

rce opposed the motion of the notebook?
Physics
2 answers:
Ratling [72]3 years ago
8 0
A frictional force (friction); which is a non-conservative force, as it dissipates the kinetic energy of the notebook through heat or/and deformation. 
Assoli18 [71]3 years ago
4 0

Answer:

The force of friction opposed the motion of the notebook.

Explanation:

Plato answer

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A cubic box of volume 5.1 x 10 -2m3 is
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Answer:

21544 N

Explanation:

Let the atmospheric pressure be 101325 Pa

The side length of the cubic box:

V = d^3 = 0.051m^3

d = \sqrt[3]{V} = \sqrt[3]{0.051} = 0.371 m

The area of the cubic box:

A = d^2 = 0.371^2 = 0.1375 m^2

20 C = 20 + 273 = 293 K

180 C = 180 + 273 = 453 K

As the volume of the air inside the closed cube is not changed, assume the ideal gas law we have

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Where P1 = 101325 Pa and T1 = 293K are the original atmospheric pressure and temperature. P2 and T2 = 453 are the new pressure and temperature after the cube gets heat up

P_2 = P_1\frac{T_2}{T_1} = 101325\frac{453}{293} = 156656 Pa

The net force on each side of the box it its pressure times side area

F = P_2A = 156656 * 0.1375 = 21544 N

8 0
3 years ago
A frictionless piston–cylinder device contains 5 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
Arte-miy333 [17]

Answer:

The work input during this process is -742 kJ

Explanation:

Given;

Initial temperature of nitrogen T₁ = 250 K

final temperature of nitrogen T₂ = 450 K

mass of nitrogen, m = 5 kg

PV^{1.4} = constant

The work input during the process is calculated as;

W = \frac{m*R(T_2-T_1)}{1-n}

where;

R is gas constant = 0.2968 kJ/kgK

substitute given values in above equation.

W = \frac{m*R(T_2-T_1)}{1-n} = \frac{5*0.2968(450-250)}{1-1.4} = -742 \ kJ

Therefore, the work input during this process is -742 kJ

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