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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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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
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1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

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the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

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  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

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