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Tanya [424]
2 years ago
11

Frank wrote several statements to summarize the relationship between the first and second laws of thermodynamics. 1 - Thermal en

ergy is conserved. 2 - Some thermal energy is lost and some is used to do work. 3 - Thermal energy flows from areas of higher to lower temperatures. 4 - Thermal energy can be transferred and transformed. Which statement relates most directly to the second law?
Physics
2 answers:
polet [3.4K]2 years ago
5 0
According to the second law, heat which is also referred to as thermal energy can be completely converted into work. 
The second statement is much related directly to the second law.
We can say that some energy is lost while other is used to do work.
frutty [35]2 years ago
5 0

Answer:

Some thermal energy is lost and some is used to do work

Explanation:

The first law of thermodynamics sates that the energy is neither created nor destroyed. This law basically states the conservation of energy.

Mathematically, it can be written as :

\Delta U=Q-W

Where,

\Delta U is the change in internal energy

Q is the heat absorbed

W is the work done

Second law states that for an isolated system the entropy should increase or remains the same. It will never decrease.

The relationship between the first and the second law is  (2) " Some thermal energy is lost and some is used to do work ".

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Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com
Lubov Fominskaja [6]

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm   ==>   area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

7 0
1 year ago
explain why the impact of one heavy stone would produce waves with higher amplitude than the impact of the light stone would
DerKrebs [107]
The heavy stone would produce waves with a higher amplitude, rather than the smaller stone, because since the stone is heaver its going to have a grater impact and displace more water to create a bigger wave.
5 0
2 years ago
Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
makvit [3.9K]
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
5 0
2 years ago
A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved
MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s

Now we can solve the problem by applying the suvat equation:

v_y^2-u_y^2=2as

where

v_y is the vertical velocity when the pea hits the ceiling

a=g=-9.8 m/s^2 is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for v_y,

v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s

Therefore, the speed of the pea when it hits the ceiling is

v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s

5 0
2 years ago
Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose
dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by h=7\ m

Suppose the dolphin jump with an initial velocity of u

so u is given by u^2=2\cdot g\cdot h

If dolphin launches at an angle \theta then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

R=\frac{u^2\sin 2\theta }{g}

substitute the value of u^2

R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}

R=2h\sin 2\theta

Range will be maximum for \theta =45^{\circ}

thus R_{max}=2\times 7\times 1=14\ m

                                     

3 0
2 years ago
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