All categories

3 years ago

Q1) An action exerted on an object which may change the object's

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

4 0

Answer: force...

Explanation:

You might be interested in

A well-insulated cup of water is too hot to make tea, so you mixed it with an equal amount of cool water to lower the temperatur

KonstantinChe [14]

Answer:The change in entropy of the total amount of water is negative as a result of the mixing.

Explanation:If you increase temperature, you increase entropy

Also More energy gives you greater entropy and randomness of the atoms.

4 0

3 years ago

A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin

lora16 [44]

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

5 0

3 years ago

If an object starts at rest and moves 60 meters north along a straight line in 2 seconds, what is the average velocity?

Lana71 [14]

Velocity: 30 m/s north

7 0

3 years ago

Read 2 more answers

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte

Lelechka [254]

Answer:

$2.25\mu C$

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

$Q=9.0 \mu C=9.0\cdot 10^{-6}C$ charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

$C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F$

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

$Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C$

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from $9.0 \mu C$ to $11.25 \mu C$ . Therefore, the additional charge that moved to the positive plate is