If Ori gives a friend three reasons for preferring soccer to basketball, that is an algorithm.

Suppose we have two negative charges, one located at the origin and carrying charge −9e, and the other located on the positive x

Elanso [62]

Answer:

It would have to be in between the two charges and be positive for force on 9e to be zero

k 9e q/x^2= k 36e 9e /d^2

q/x^2 = 36e/d^2

q = 36e x^2/d^2

Now we need force on 36e to be zero

k 36e q/(d-x)^2 = k 36e 9e/d^2

q/(d-x)^2 = 9e/d^2

36 x^2/d^2 = 9/d^2 (d-x)^2

36 x^2 = 9 (d-x)^2

x=d/3

q = 36e (d/3)^2/d^2 = 36e/3^2 = 4 e

7 0

4 years ago

30 points have a good day

lakkis [162]

Answer:

thank you :)

Explanation:

8 0

3 years ago

Consider a drainage basin having 60% soil group A and 40% soil group B. Five years ago the land use pattern in the basin was ½ w

victus00 [196]

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf

3 0

4 years ago

A global resources company uses data-intensive, cloud-based simulation software, but users in remote locations find that the res

lakkis [162]

The type of technology illustrated in the given example is called; Edge Computing

We are told that;

- Company uses data-intensive, cloud-based simulation software

- Company decides to deploy multiple instances of the application in locations closer to the end users due to the fact that users in remote locations complained of lack of responsiveness.

Now, the type of technology illustrated from the solution the company is trying to perform is called Edge Computing.

This is because Edge Computing fundamentally is a type of technology that is used to bring computation and data storage closer to the user devices that collects them instead of depending on a central location that may be thousands of kilometers away.

Now, the edge computing technology is usually done to ensure that real-time data does not suffer from latency issues that can affect the performance of applications.

Read more on edge computing at;brainly.com/question/23858023

6 0

2 years ago

A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures

Alex777 [14]

Answer:

a) T_5 = 782.8 K

b) W_cyc = 108.04 KJ/kg

c) n_th = 22.47 %

d) X_dest = 289.924 KJ/kg , X_exhaust = 126.6768 KJ/kg

Explanation:

Given:

- P_2 / P_1 = 7

- T_4 = 1150 K

- T_1 = 310 K

- n_s,comp = 0.75

- n_s,turb = 0.82

- R_air = 0.287 KJ/kg

Find:

- T_5 - Temperature of air at turbine exit ?

- W_cycle ?

- n_th ?

- X_dest , X_exhaust ?

Solution:

Assumptions:

1) The cycle operates at steady-state.

2) Air is the working fluid and it behaves as an ideal gas.

3) The Brayton Cycle is modeled as as a closed cycle.

4) The combustor is replaced by a HEX. (External Combustion)

5) The compressor and turbine are not internally reversible.

6) Changes in kinetic and potential energies are negligible.

7) Air has variable specific heats.

8) The compressor and turbine are adiabatic.

Analysis:

- The efficiency of turbine is given by:

n_s,turb = (H_4 - H_5) / (H_4 - H_5,s)

- For H_4 and S_ 4 we have T_4 = 1150 K, use the ideal gas air property table:

T_4 = 1150 K -------------> S_4 = 3.129 KJ/kgK , H_4 = 1219.25 KJ/kg

- For H_5s, the enthalpy of the effluent from a hypothetical isentropic turbine. We can do this because we know the values of two intensive variables: P_5 and S_5 = S_4. The key to using this information is the 2nd Gibbs equation:

S_5,s,o = S_4,s,o + R_air*Ln(P_5 / P_4)

S_5,s,o = 3.129 + 0.287*Ln(1 / 7) = 2.57054 KJ/kgK

- Now use the value S_5,s,o and ideal gas air property table and evaluate:

S_5,s,o = 2.57054 KJ/kgK ---------> T_5,s = 698.6 K ,

H_5s = 711.72 KJ/kg

- Now use the efficiency relation for turbine:

H_5 = H_4 - n_s,turb*(H_4 - H_5,s)

H_5 = 1219.25 - 0.82*(1219.25-711.72)

H_5 = 803.08 KJ/kg

- Using H_5 and ideal gas air property table and evaluate:

H_5 = 803.08 KJ/kg -----------> T_5 = 782.8 K , S_5 = 2.6940 KJ/kg-K

- In order to determine the specific shaft work for the cycle, we need to determine the specific shaft work for the compressor and for the turbine

W_cyc = W_comp + W_turb

W_cyc = H_1 - H_2 + H_4 - H_5

- The efficiency of compressor is given by:

n_s,comp = (H_1 - H_2,s) / (H_1 - H_2)

- For H_1 and S_ 1 we have T_1 = 310 K, use the ideal gas air property table:

T_1 = 310 K -------------> S_1 = 1.73498 KJ/kgK , H_1 = 310.24 KJ/kg

- For H_2s, the enthalpy of the effluent from a hypothetical isentropic compressor. We can do this because we know the values of two intensive variables: P_2 and S_2 = S_1. The key to using this information is the 2nd Gibbs equation:

S_2,s,o = S_1,s,o + R_air*Ln(P_2 / P_1)

S_2,s,o = 1.73498 + 0.287*Ln(7) = 2.29343 KJ/kgK

- Now use the value S_2,s,o and ideal gas air property table and evaluate:

S_2,s,o = 2.29343 KJ/kgK ---------> T_2,s = 537.1 K ,

H_2s = 541.34 KJ/kg

- Now use the efficiency relation for compressor:

H_2 = H_1 - (H_1 - H_2,s)/n_comp

H_2 = 310.24 - (310.24-541.34)/0.72

H_2 = 618.37 KJ/kg

Hence,

The work out for the cycle is:

W_cyc = 310.24 - 618.37 + 1219.25 - 803.08

W_cyc = 108.04 KJ/kg

- The thermal efficiency of a cycle is:

n_th = W_cyc / Q_H

Q_H = H_4 - H_3

- The effectiveness of re-generator is e:

e = (H_3 - H_2) / (H_5 - H_2)

H_3 = (H_5 - H_2)*e + H_2

H_3 = (803.08 - 618.37)*0.7 + 618.37 = 738.43 KJ/kg

Hence,

Q_H = 1219.25 - 738.43 = 480.81 KJ/kg

Finally,

n_th = 108.04 / 480.81 = 22.47 %

- The amount of heat loss is given by:

Q_L = H_6 - H_1

H_6 = H_5 + H_2 - H_3 = 803.08 + 618.37 - 738.43 = 682.97 KJ/kg

Q_L = 682.97 - 310.24 = 372.73 KJ/kg

- The amount of exergy destroyed for whole cycle:

X_dest = T_L * ( Q_L / T_L - Q_H / T_H)

X_dest = 310 * (372.73 / 310 - 480.81 / 1800)

X_dest = 289.924 KJ/kg

- The amount of exergy of exhaust gasses:

X_exhaust = H_6 - H_0 - T_L*(S_6 - S_o )

X_exhaust = 682.97 - 310.24 - 310*(2.52861 - 1.73489 )

X_exhaust = 126.6768 KJ/kg

4 0

3 years ago