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3 years ago

Agricultural economics is a study of how agriculture and business are related. False True

1 answer:

8 0

Answer:

False

Explanation:

Agricultural economics concerns itself with the study of the production and consumption of food in both developed and developing countries along with analysis of the policies that shape the world's largest industry.

I think this is right but my apologies if it’s not

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In a steady flow device, the properties of the system remains constant with time. a)True b) False

Leviafan [203]

Answer:

True

Explanation:

By definition of steady flow we have

$\frac{\partial f(x,y,z,t) }{\partial t}=0$

where f(x,y,z,t) is any property of the system under consideration

=> f(x,y,z,t) = constant

7 0

3 years ago

Two Technicians are discussing ShopKey Pro. Technician

Leno4ka [110]

Answer:

Technician B

Explanation:

i took he test already

6 0

3 years ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

2 years ago

What are the controlling LRFD load combinations for dead and floor live load?

yuradex [85]

Answer:

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

Explanation:

Load and Resistance Factor Design

there are 7 basic load combination of LRFD that is

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

and

here load factor for L given ( * ) mean it is permitted = 0.5 for occupancies when live load is less than or equal to 100 psf

here

D is dead load and L is live load

E is earth quake load and S is snow load

W is wind load and R is rain load

Lr is roof live load

3 0

3 years ago

In RSA Digital Signature, Suppose Bob wants to send a signed message (x = 4) to Alice. The first steps are exactly t eps are exa

Luda [366]

Answer:

what r u on

Explanation:

4 0

3 years ago