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Stells [14]
1 year ago
11

The linear expansion coefficient of aluminium is 24 × 10^-6 (°C)^-1. When thetemperature is 33 °C, a spherical aluminium ball ha

s a diameter of 11.21 cm.A) Calculate the diameter (in cm) of the aluminium ball when the temperature is raisedto 257° C? (Round your answer to 3 decimal places.)B) Calculate the change in temperature (in °C) needed to increase the volume of thealuminium ball by 3.24%.
Physics
1 answer:
KatRina [158]1 year ago
6 0

A) Use the following formula for the linear expansion, in this case, of the diameter of the spherical ball:

d=d_o\lbrack1+\alpha(T_2-T_1)\rbrack

where,

do: initial diameter = 11.21cm

d: final diameter = ?

α: linear expansion coefficient = 24*10^6 /°C

T2: final temperature = 257°C

T1: intial temperature = 33°C

Replace the previous values of the parameters into the formula for d and simplify:

Hence, the diameter of the spherical ball is 11.27cm

B) First, consider what is the change in the radius of the sphere, as follow:

\Delta V=0.0324V_o

Now, use the following formula for the change in volume with the temperature:

\Delta V=\beta V_o\Delta T=3\alpha V_o\Delta T

where you have used the equivalence β = 3α.

Solve the equation above for the change in temperature and replace the values of the other parameters:

\Delta T=\frac{\Delta V}{3\alpha V_o}=\frac{0.0324V_o}{3(24\cdot10^{-6}(\degree C)^{-1})V}=450\degree C

Hence, the change in temperature is 450°C

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Answer:

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(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

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