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fredd [130]
4 years ago
5

A pesky rabbit has been feeding on Mrs. Cromwell’s prized flowers. In order to put an end to this she devised the simple trap be

low to catch the rabbit. The crate weighs 13lb and is supported by stick BC. Find the minimum diameter of stick BC if the normal stress in it is limited to 100 psi.

Physics
1 answer:
Ivahew [28]4 years ago
4 0

Answer:

The minimum diameter of the stick to bear the normal stress as 100 psi is 0.1785 in.

Explanation:

Taking moment along point A

\sum M_A=0\\F_{BC} sin 60 \times 48 -13 \times 8=0\\F_{BC}=2.502 lb_f

Also normal stress is given as 100 psi

Now

                      \sigma=\frac{F}{A}\\100=\frac{2.502}{\pi \frac{d^2}{4}}\\d^2 =\frac{4 \times 2.502}{100 \pi }\\d^2=0.031856\\d=\sqrt{0.031856}\\d=0.1785 inch

The minimum diameter of the stick to bear the normal stress as 100 psi is 0.1785 in.

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A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie
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a) 3.27\cdot 10^{-3} J

b) 11.60\cdot 10^{-3} J

c) 8.33\cdot 10^{-3} J

Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

C=13.5\cdot 10^{-6}F is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

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