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vivado [14]
4 years ago
6

Under which condition is no work done?

Physics
2 answers:
REY [17]4 years ago
8 0

Answer: It's D

Explanation: just took the test

koban [17]3 years ago
5 0

Answer:

pulls object

Explanation:

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You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
3 years ago
A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. Ho
natita [175]

Answer:

Use Fc centripetal force as positive and W the weight as negative

N = m v^2 / R + m g

v^2 = (N - m g) R / m

v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2

v = 18.1 m/s

Note: N - m g is the net force producing the centripetal force

5 0
2 years ago
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system an
Trava [24]

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

8 0
3 years ago
Write a formular to show force experienced by charge in magnetic field.​
Mars2501 [29]

Answer:

{ \boxed{ \sf{F = Bev}}}

Explanation:

F » Force

B » Magnetic flux density

e » magnitude of charge

v » voltage

7 0
3 years ago
a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoi
Free_Kalibri [48]

Answer:

Q₂ = 5833.33 J

Explanation:

First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:

η = W/Q₁

where,

η = efficiency of engine = 30% = 0.3

W = Work done by engine = 2500 J

Q₁ = Heat supplied to the engine = ?

Therefore,

0.3 = 2500 J/Q₁

Q₁ = 2500 J/0.3

Q₁ = 8333.33 J

Now, we find the heat discharged to lower temperature reservoir by using the formula of work:

W = Q₁ - Q₂

Q₂ = Q₁ - W

where,

Q₂ = Heat discharged to the lower temperature reservoir = ?

Therefore,

Q₂ = 8333.33 J - 2500 J

<u>Q₂ = 5833.33 J</u>

7 0
3 years ago
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