Answer:
a.) The main scale reading is 10.2cm
b.) Division 7 = 0.07
c.) 10.27 cm
d.) 10.31 cm
e.) 10.24 cm
Explanation:
The figure depicts a vernier caliper readings
a.) The main scale reading is 10.2 cm
The reading before the vernier scale
b.) Division 7 = 0.07
the point where the main scale and vernier scale meet
c.) The observed readings is
10.2 + 0.07 = 10.27 cm
d.) If the instrument has a positive zero error of 4 division
correct reading = 10.27 + 0.04 = 10.31cm
e.) If the instrument has a negative zero error of 3 division
correct reading = 10.27 - 0.03 = 10.24cm
155Ω
Explanation:
R = R ref ( 1 + ∝ ( T - Tref)
where R = conduction resistance at temperature T
R ref = conductor resistance at reference temperature
∝ = temperature coefficient of resistance for conductor
T = conduction temperature in degrees Celsius
T ref = reference temperature that ∝ is specified at for the conductor material
T = 600 k - 273 k = 327 °C
Tref = 300 - 273 K = 27 °C
R = 50 Ω ( 1 + 0.007 ( 327 - 27) )
R = 155Ω
The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.
<h3>Pressure and temperature at equilibrium </h3>
The relationship between pressure and temperature can be used to determine the height risen by the water.

where;
- V₁ = AL
- V₂ = A(L - y)
- P₁ = Pa
- P₂ = Pa + ρgh
- T₁ = 20⁰C = 293 K
- T₂ = 10⁰ C = 283 k

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.
The complete question is below:
A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?
Learn more about thermal equilibrium here: brainly.com/question/9459470
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Answer:
x=0.154kg
Explanation:
(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0
(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0
Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J