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Orlov [11]
3 years ago
8

A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s

hear stress is not to exceed 25 MPa?
Physics
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder =  150 mm

thickness of cylinder = 6 mm

maximum shear stress =  25 MPa

we know that

hoop stress is given as =\frac{pd}{2t}

axial stress is given as =\frac{pd}{4t}

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}

25 = \frac{pd}{8t}

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

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Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
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snow_tiger [21]

Answer:

It's B, anything about a circle is Stationary

4 0
3 years ago
two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist
Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

R_{eq}=R_1+R_2

For parallel combination,

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

When 6 ohm and 3 ohm are in series,

R_s=6+3\\\\R_s=9\ \Omega

When 6 ohm and 3 ohm are in paralle,

\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0
2 years ago
Town A lies 15 km north of town B. Town C lies 10 km west of town A. A small plane flies directly from town B to town C. What is
ZanzabumX [31]

Answer:

the correct answer is b

Explanation:

5 0
3 years ago
Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5
pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

\dfrac{F}{L}= \dfrac{KI_1I_2}{d}

I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}

I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}

    I₂ = 32.67 A

distance where the magnetic field is zero

\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}

y_1 = 0.248\ m

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0
3 years ago
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