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hjlf
3 years ago
13

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Engineering
1 answer:
Andreyy893 years ago
4 0
I don’t know but good luck
You might be interested in
the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i
Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

V(t) = 5(1-e^{-0.5t})

The current flowing through the inductor is given by

i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\

i(t) = 1000t +2000e^{-0.5t} -2000\\

So the current at t = 2 seconds is

i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A

The energy stored in the inductor at t = 2 seconds is

W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J

4 0
2 years ago
Which utility program reads an assembly language source file and produces an object file?
iragen [17]

Answer:

Assembler

Explanation:

An assembler can be define as a computer utility program that read, interpret and convert software programs written in low level assembly language into an object file, machine language, code and instruction that can be understood and executed by a computer.

5 0
3 years ago
Is there a way to get the answers to a NCCER book test?
sergeinik [125]

Answer:

go on google and type NCEER book answers

3 0
3 years ago
A driver is traveling at 90 km/h down a 3% grade on good, wet pavement. An accident
Paul [167]

Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ -  μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction  coefficient is 0.35

4 0
3 years ago
7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute
Crank

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

<h3>How to draw Shear Force and Bending Moment Diagram?</h3>

A) We can see the beam loaded in the first image attached.

For the shear diagram, let us calculate the shear from point load to point load.

From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0

V = -20 KN

From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0

V = 28 KN

From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0

V = 8 KN

From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0

V = -12 KN

For the bending moment diagram, let us calculate the bending moment from point load to point load.

At point A, the bending moment would be zero. Thus, M_A = 0 KN.m

At point C, taking moment about point C and equating to zero gives;

M_C = 0. Thus; 20(0.225) + M = 0

M = -4.5 KN.m

At point D, taking moment about point D and equating to zero gives;

M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0

M = 3.9 KN.m

At point E, taking moment about point E and equating to zero gives;

M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0

M = 5.7 KN.m

At point B, taking moment about point E and equating to zero gives;

M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0

M = 2.1 KN.m

2) From the attached diagrams, we can deduce that;

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

Read more about shear force & bending moment diagram at; brainly.com/question/14834487

#SPJ1

4 0
2 years ago
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