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marishachu [46]
3 years ago
8

What is meant by physical balance? ​

Physics
1 answer:
serious [3.7K]3 years ago
8 0

Answer:

it is a balance in measuring weights in milligram

Explanation:

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Answer: I actually need the same answer

Explanation:

5 0
3 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
What component of an atom determines what element it is?
serg [7]
The number of protons in the element.
8 0
4 years ago
The process by which plastic deformation is produced by dislocation motion is called
QveST [7]

Answer:gock uwu

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3 0
3 years ago
How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules
dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
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