All categories

3 years ago

What occurs at thermal equilibrium?

Physics

2 answers:

Ainat [17]3 years ago

8 0

Answer:

C, JUST TOOK

Explanation:

Gnesinka [82]3 years ago

6 0

C. Equilibrium means at an equal or same state; the two substances have reached thermal equilibrium, so they are at an equal state of temperature. No energy is transferred.

You might be interested in

3. A tennis ball is thrown straight up into the air with an initial velocity of 8.9 m/s. -9.81 gravity

otez555 [7]

Answer:

a. time is 0.91s

b. height is 4.04m

Explanation:

t = u/g
t = 8.9/9.81 = 0.91s
the time it takes for the tennis ball to reach its highest point is 0.91 seconds

S = u²/2g
S = 8.9²/2x9.81
s = 79.21/19.62
s = 4.04m

5 0

3 years ago

Why does an ice cube melt when you hold it in your hand? Question 1 options: Heat from the ice cube is transferred to your hand.

Sunny_sXe [5.5K]

Your hand transferred temperature to he ice cube

8 0

3 years ago

Read 2 more answers

Let σa, σb, σc, and σd be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge densities to b

aniked [119]

Answer:

a) σa−σb−σc−σd=0

Explanation:

The parallel plate capacitor is the one in which two metal plates are connected in parallel with some distancing among them. The electric field from both plates is denoted by E = σ / 2ϵ0. The σ is the charge density. The Electric field in plate I will vanish when the surface charge of σa is positive and rest of the charges are negative. The correct option is a.

7 0

3 years ago

First sign of lung cancer

quester [9]

the most common first sign of lung cancer is a constant cough. there could also be coughing up blood or reddish colored spit, and chest pain when you laugh, cough, and breathe.

5 0

3 years ago

Read 2 more answers

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago