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sergeinik [125]

2 years ago

7

Compound axial member ABC has a uniform diameter of d = 1.50 in. Segment (1) is an aluminum [E1 = 10,000 ksi] alloy rod with a l

ength of L1 = 90 in. Segment (2) is a copper [E2 = 17,000 ksi] alloy rod with a length of L2 = 130 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 2,100 in./in. in the longitudinal direction. What is the total elongation of member ABC?

1 answer:

skad [1K]2 years ago

5 0

Explanation:

Below is an attachment containing the solution.

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Answer:

the speed is equal to 6 m/s

Explanation:

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2 years ago

A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3

Klio2033 [76]

Answer:

$\theta=30^{\circ}$

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

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Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

$F=ILB\ sin\theta$

$sin\theta=\dfrac{F}{ILB}$

$sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}$

$\theta=30^{\circ}$

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0

2 years ago

A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

$H = \frac{nI}{l}$

Here,

n = Number of turns of the coil

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l = Length of the coil

From the above equation, the number of turns of the coil is,

$n = \frac{Hl}{I}$

The magnetic field is again given by,

$H = \frac{B}{\mu_t}$

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

$n = \dfrac{\frac{B}{\mu_t}l}{I}$

Replacing with our values we have that,

$n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}$

$n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}$

$n = 27.5 \approx 28$

Therefore the number of turn required is 28Truns

4 0

2 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

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(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

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As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

2 years ago

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ololo11 [35]

Answer:

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Explanation:

4 0

2 years ago