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Svet_ta [14]
1 year ago
14

Rosa pours a cup of boiling water into a pot of room-temperature water. According to the second law of thermodynamics, what will

occur
Physics
1 answer:
11111nata11111 [884]1 year ago
8 0

<u>Thermal energy</u><u> from the room-temperature water will continuously flow to the boiling water.</u>

  • The second law states, in a straightforward manner, that heat cannot naturally go "uphill."
  • When a pan of boiling water and a pan of ice are in touch, the hot water cools and the ice melts and warms up.

<h3>THE FIRST LAW OF THERMODYNAMICS</h3>
  • Adiabatic Process - is a procedure that is carried out without the system's heat content changing.
  • Water is heated to a temperature of 1000C during the boiling process, making it an isothermal process. As steam, the excess heat leaves the system.

Learn more about first law of thermodynamics brainly.com/question/3808473

#SPJ4

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Read 2 more answers
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
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