A steel railroad track has a length of 40 m when the temperature is −5 ◦C. What is the increase in the length of the rail on a hot day when the temperature is 35 ◦C? The linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1 . Answer in units of m.

ΔL = 0.0176m

Explanation:

We are given;

Length of railroad track; L = 40 m

First Temperature; T1 = −5 ◦C

Second temperature; T2 = 35 ◦C

linear expansion coefficient of steel; α = 11 × 10^(−6) (◦C)^(-1)

The increase in length is given by the equation;

ΔL = α•L•ΔT

Where,

α is linear coefficient

L is length

ΔT is change in temperature.

ΔT = first temperature - Second Temperature

Thus, ΔT = 35 - (-5) = 35 + 5 = 40°C

Thus,plugging in relevant values,

ΔL = α•L•ΔT = 11 × 10^(−6)•40•40

ΔL = 0.0176m

0.12 C.

Explanation:

The following data were obtained from the question:

Capacitor (C) = 10 mF

Potential difference (V) = 12 V

Charge (Q) =?

Next, we shall convert 10 mF to farad (F). This can be obtained as follow:

1 mF = 1×10¯³ F

Therefore,

10 mF = 10 mF × 1×10¯³ F/ 1mF

10 mF = 0.01 F

Therefore, 10 mF is equivalent to 0.01 F.

Finally, we shall determine the charge on the capacitor as illustrated below:

Capacitor (C) = 0.01 F

Potential difference (V) = 12 V

Charge (Q) =?

Q = CV

Q = 0.01 × 12

Q = 0.12 C

Therefore, the charge on the capacitor is 0.12 C.