Which of the following parts tells the transmission how fast or slowly a vehicle is going? A) pump

Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r

My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0

4 years ago

Who developed the process of blueprinting?

VikaD [51]

Answer: C.) John Herschel

3 0

3 years ago

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

$\epsilon_{car}=1-\frac{T_c}{T_H}$

Where $T_c$ is the atmospheric temperature and $T_H$ is the maximum burn temperature.

For the case (B), the efficiency we will use is:

$\epsilon_{b}=0.4\epsilon_{car}$

The work done by the engine can be calculated as:

$W=\epsilon Q=\epsilon H_v\cdot m_{fuel}$ where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

$m=\frac{P\cdot t}{\epsilon H_v}$

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

$TC=m\cdot c$

8 0

3 years ago

The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both

DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

L^2 = x_a^2 + y_b^2 ....... 1

y_b = sqrt( 20^2 - 16^2 )

y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

0 = 2*x_a*v_a + 2*y_b*v_b

0 = x_a*v_a + y_b*v_b ..... 2

0 = 16*36 + 12*v_b

v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

0 = 9 + 4*a_a/3 + 16 + a_b

4*a_a/3 + a_b = -25

4*a_a + 3*a_b = -75 .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

F_x = m_a*a_a

P - R_r*16/20 = m_a*a_a

- For Slider B, Apply Newton's second law of motion in y direction:

F_y = m_b*a_b

- R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

P - 4*m_b*a_b / 3 = m_a*a_a

3P = 3*m_a*a_a + 4*m_b*a_b ... 4

- Where, P = Is the force acting on slider A

P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.

5 0

3 years ago

1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T

Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

8 0

3 years ago