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Ber [7]
3 years ago
15

How much work does the electric field do in moving a proton from a point with a potential of +V1 = +185 V to a point where it is

V2 = 55.0 V?
Engineering
1 answer:
Arada [10]3 years ago
4 0

Answer:

W=  2.08 x 10⁻¹⁷  J

Explanation:

Given that

Initial potential V₁ =185 V

Final potential V₂ = 55 V

We know that charge of the proton

q=1.6 x 10⁻¹⁹ C

Work done is given as

W= q ΔV

q=Charge

ΔV=Potential difference

W=Work done

Now by putting the values in the above equation then we get

W= 1.6 x 10⁻¹⁹  ( 185 - 55 ) J

W=208 x 10⁻¹⁹ J

W=  2.08 x 10⁻¹⁷  J

Therefore the work done will be 2.08 x 10⁻¹⁷  J.

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Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

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Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

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Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0
2 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

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