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Vedmedyk [2.9K]
3 years ago
11

Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC

Physics
1 answer:
nignag [31]3 years ago
5 0

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

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A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How mu
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Answer:

84.6 J

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W=Fd cos \theta

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Read 2 more answers
Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o
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The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

Q=ne

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C

The potential difference V between the disks separated by a distance d is given by,

V=Ed

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V

The capacitance C of the capacitor is given by,

C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F

The capacitance of a parallel plate capacitor is given by,

C=\frac{\epsilon_0A}{d}

Here, ε₀ is the permittivity of free space  and A is the area of the disks.

Rewrite the expression for A.

C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3

the area A of the disks is given by,

A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }

Here, D is the diameter of the disk.

D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm

The diameter of each disc is found to be 1.32 cm.



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