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OverLord2011 [107]
3 years ago
6

1. Represent each of the following combinations of units in the correct SI form using the appropriate prefix: (a) μMN, (b) N/μm,

(c) MN/ks2, and (d) kN/ms.
2. Represent each of the following quantities in the correct SI form using the appropriate prefix: (a) 0.000431 kg, (b) 35.3(103) N, (c) 0.00532 km.
3. Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/nm, and (c) mN/(kg⋅μs).
Engineering
1 answer:
Burka [1]3 years ago
6 0

Answer:

Kindly check explanation

Explanation:

(a) μMN, (b) N/μm, (c) MN/ks2, and (d) kN/ms.

(a) μMN = (10^-6 * 10^6) = 10^(-6 + 6) = 10^0 = N

b) N/μm = N / 10^-6 m = 10^6 * N/m = MN/m

(c) MN/ks2 = 10^6N / (10^3 s)^2

10^6 N / 10^6s^2 = 10^6 * 10^-6 N /s^2 = N/s^2

D) kN/ms = 10^3N / 10^-3 s = 10^3 * 10^3 * N/s = 10^6N / s = MN/s

2)

a) 0.000431kg = 431 × 10^6 kg = 431 * 10^9g = 431Gg

b) 35.3 × 10^3 N

10^3 = kilo(K)

35.3 KN

C) 0.00532km = 5.32 * 10^3 km = 5.32 * 10^3 * 10^3 = 5.32 * 10^6 = 5.32Mm

3) Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/nm, and (c) mN/(kg⋅μs).

a) Mg/ms = 10^6g / 10^-3s = 10^6 * 10^3 g/s = 10^9 g/s = Gg/s

b) N/nm = N / 10^-9 m = 10^9 N/m = GN/m

c) mN/(kg⋅μs) = 10^6N / kg(10^-6s) = 10^12N/(kg.s)

= TN/(kg.s)

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Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
3 years ago
A farmer has 12 hectares of land on which he grows corn, wheat, and soybeans. It costs $4500 per hectare to grow corn, $6000 to
maw [93]

The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans

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He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;

C + W + S = 12   ----(1)

It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;

45C + 60W + 50S = 63750  -----(2)

He will grow twice as many hectares of wheat as corn. Thus;

W = 2C    ------(3)

Put 2C for W in eq 1 and eq 2 to get;

C + 2C + S = 1200

3C + S = 1200     -----(4)

45C + 60(2C) + 50S = 63750

45C + 120C + 50S = 63750

165C + 50S = 63750    ------(5)

Solving eq 4 and 5 simultaneosly gives;

C = 250 and W = 500

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5 0
2 years ago
Select the statement that is false.
ra1l [238]

Answer:

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8 0
3 years ago
‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

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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
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