Find the resistance of a circuit that draws 4 amperes with 8 volts applied?

As you push a toggle bolt into a wall, the

MakcuM [25]

Answer:

Explanation:
One the object is attached the toggle it's placed onto the bolt. The wings should be open in the direction of the bolt head. The toggle is than collapsed and the entire toggle bolt is inserted into the hole in the wall. Once the wings are fully through the opening they will spring open on the other side.
Hopt it's help thanks..

5 0

2 years ago

The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh

lapo4ka [179]

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, $N_{A}$ = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Now we know that for maximum velocity,

$\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }$

$\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }$

$N_{B}$ = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Therefore moment of inertia of flywheel, I = m. $k^{2}$

=30 X $0.1^{2}$

= 0.3 kg- $m^{2}$

Now torque on the output shaft

T₂ = I x ω

= 0.3 X 1064.2 rpm

= $0.3\times \frac{2\pi \times 1064.1}{60}$

= 33.429 N-m

Torque on the Shaft B is 33.429 N-m

4 0

3 years ago

A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o

Norma-Jean [14]

Answer:

q=2313.04 $W/m^2$

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

$R=\dfrac{t}{KA}$

These are connect in series

$R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}$

$R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W$

$R=\dfrac{23}{56A}\ K/W$

Heat transfer

$Q=\dfrac{\Delta T}{R}$

$Q=56A\times \dfrac{980-30}{23}\ W$

So heat flux

q=2313.04 $W/m^2$

Lets temperature between interface is T

Now by equating heat in both bricks

$\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}$

So T=690.86°C

6 0

3 years ago

Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr

omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

Q_in = Δ u_2,3

q_2,3 = u_3 - u_2

q_2,3 = c_v*(T_H - T_2)

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

q_2,3 = c_v*(T_H - T_L*r^(n-1) )

q_2,3 = 0.718*(1173-288*8(1.3-1) )

q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n + q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

Q_out = Δ u_4,1

q_4,1 = u_4 - u_1

q_4,1 = c_v*(T_4 - T_L)

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

q_4,1 = c_v*(T_H*r^(1-n) - T_L )

q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

q_net,in = q_3,4+q_2,3

q_net,in = 129.8+456

q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

q_net,out = q_4,1+q_1,2

q_net,out = 244+60

q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

n = 1 - q_net,out / q_net,in

n = 1 - 304/585.8

n = 0.481

6 0

3 years ago

The electron concentration in silicon at T = 300 K is given by

puteri [66]

Answer:

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

Explanation:

From the question we are told that:

Temperature of silicon $T=300k$

Electron concentration $n(x)=10^{16}\exp (\frac{-x}{18})$

$\frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})$

Electron diffusion coefficient is $Dn = 25cm^2/s \approx 2.5*10^{-3}$

Electron mobility is $\mu n = 960 cm^2/V-s \approx0.096m/V$

Electron current density $Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2$

Generally the equation for the semiconductor is mathematically given by

$Jn=qb_n\frac{dn}{dx}+nq \mu E$

Therefore

$-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E$

$E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}$

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

7 0

2 years ago