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4 years ago

Convert mechanical energy into electric energy. What can he use?

Engineering

1 answer:

Nina [5.8K]4 years ago

7 0

Answer:

<h2>Generator </h2>

Explanation:

A generator converts mechanical energy into electrical energy

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Are there engineering students here?

leva [86]

Uh, I’d assume so because Brainly has a whole section of questions for them.

7 0

3 years ago

Read 2 more answers

Q1: The first option should always be to get out safely (RUN)

nekit [7.7K]

Answer:

Q1 true

Q2 true

And other I am confuse

6 0

3 years ago

A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f

jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0

3 years ago

Explain the working of McLeod gauge for measurement of very low pressure.

AfilCa [17]

Explanation:

McLeod gauge:

It is used to measure very low pressure of gas.It measure gas pressure by the help of mercury(Hg).it measure absolute pressure of gas.

McLeod gauge on the principle of Boyle's law.Boyle's law state that ,at constant temperature the pressure of gas is directly proportional to its volume.From Boyle's law

$P_1V_1=P_2V_2$

A sample of gas is taken from vacuum then its pressure is measured by help Hg.

4 0

3 years ago

. Two rods, with masses MA and MB having a coefficient of restitution, e, move

GarryVolchara [31]

Answer:

a) $V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$

b) $U_A = 3.66 m/s$

$V_B = 4.32 m/s$

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let $U_A$ be the initial velocity of rod A

Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

Using the principle of conservation of momentum:

$M_AU_A + M_BU_B = M_AV_A + M_BV_B$ ............(1)

Coefficient of restitution, $e = \frac{V_B - V_A}{U_A - U_B}$

$V_A = V_B - e(U_A - U_B)$ ........................(2)

Substitute equation (2) into equation (1)

$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

Solving for $V_B$ in equation (3) above:

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

Rod A is said to be at rest after the impact, $V_A = 0 m/s$

Substitute these parameters into equation (7)

$0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s$

To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

$V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s$

c) Impulse, $I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)$

$I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))$

I = 0 $kg m/s^2$

d) % $\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%$

% $\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%$

% $\triangle KE = -47.85 \%$

7 0

3 years ago