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ad-work [718]
3 years ago
8

A low-temperature physicist wishes to publish his experimental result that the heat capacity of a nonmagnetic dielectric materia

l between 0.05K and 0.5K varies as
A√ T + BT^3

As editor of the journaL should you accept the paper for publication?
Engineering
1 answer:
tatiyna3 years ago
6 0

Answer:

Yes, this because the specific heat does not violate the third law of thermodynamics.

Explanation:

The third law of thermodynamic is usually used for a closed system to relate the thermodynamic properties of the system at equilibrium conditions. For this law, a body that exists at a temperature of 0 K will cease to move or stop moving. It can be inferred that heat capacity at (T = 0 K) is equivalent to 0. Therefore, the equation is valid for the given temperature range.  

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Technician A says that the paper test could detect a burned valve. Technician B says that a grayish white stain on the engine co
Bumek [7]

Answer:

Both Technician A and B are correct.

Explanation: Both are correct, but keep note that different color coolants leave different color  stains.

6 0
3 years ago
Is normally a large red cable connected to the battery
Igoryamba

Answer:

yes

Explanation:

its the ground cable

4 0
3 years ago
Read 2 more answers
Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1
steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

3 0
3 years ago
Which of the following explains the main reason to cut a piece of wood on the outside of the measurement mark?
maks197457 [2]
I think it’s D ?? I’m not completely sure tho
4 0
3 years ago
You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from
Aleksandr [31]

Answer:

a) \frac{Ws}{Es}  = \frac{200}{1+1.2s}

b) attached below

c) type zero system

d) k > \frac{g}{200}

e) The gain K increases above % error as the  steady state speed increases

Explanation:

Given data:

Motor voltage  = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; \frac{K1}{1+St} be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= \frac{Ws}{Es}  = \frac{200}{1+1.2s}

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > \frac{g}{200}

7 0
3 years ago
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