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hram777 [196]
4 years ago
9

A 750-turn solenoid, 24 cm long, has a diameter of 2.3 cm . A 19-turn coil is wound tightly around the center of the solenoid. P

art A If the current in the solenoid increases uniformly from 0 to 5.1 A in 0.70 s , what will be the induced emf in the short coil during this time? Express your answer to two significant figures and include the appropriate units. |E| = nothing nothing Request Answer Provide Feedback
Engineering
1 answer:
oksian1 [2.3K]4 years ago
6 0

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

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A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T =  1.125 × 10⁶ / 75

T = 15000 N

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7 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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tatuchka [14]

Answer:

B. to lock the tape into place

Explanation:

the button on the front of the housing locks the tape into place when pressed, preventing the tape from being pulled out further it retracting

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