1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
muminat
3 years ago
15

A non-entrepreneurship, work-based, agricultural type of SAE, in which a student learns and gains skills in a paid or unpaid pos

ition describe which type of SAE?
Engineering
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

Placement.

Explanation:

SAE is an acronym for Supervised Agricultural Experience programs and it is typically a planned, practical-based program designed to help students develop competent skills and experience in their various career choice. Basically, it enhances the academic knowledge and agricultural skills acquired by the students in the classroom.

Generally, there are four (4) main types of Supervised Agricultural Experience Programs;

1. Entrepreneurship SAE.

2. Exploratory SAE.

3. Research and Experimentation SAE.

4. Placement SAE.

A non-entrepreneurship, work-based, agricultural type of Supervised Agricultural Experience program (SAE), in which a student learns and gains skills in a paid or unpaid position describe a placement.

Some examples of organizations or businesses that provide placement SAE to students includes; lawn services, veterinary clinics, soil conservation firms, livestock farms, floral shops, grain farms etc.

You might be interested in
Short-term memoryA) has a larger storage capacity than long-term memory.B) takes longer to retrieve than long-term memory.C) inv
zaharov [31]

Answer:D

Explanation:

Take longer time to retrieve than long term memory, involves transient modifications in the function of pre existing synapses, such as channel modifications.

6 0
3 years ago
In casting experiments performed using a certain alloy and type of sand mold, it took 170 sec for a cube-shaped casting to solid
poizon [28]

Answer:

Answer for the question is : Solidification time will be same i.e. 170. See attached file for explanation.

Explanation:

Download pdf
7 0
2 years ago
Read 2 more answers
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is
Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

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
Other questions:
  • Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that
    7·1 answer
  • Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!
    13·1 answer
  • Linear Time Invariant Systems For each of the systems below an input x(t) and the output y(t) are plotted. Determine whether eac
    5·1 answer
  • Use the APWA 5600 methodology to compute the 1/25 design discharge for the following catchment. The catchment area consists of 1
    9·1 answer
  • 4. A 1 m3 rigid tank has propane at 100 kPa, 300 K and connected by a valve to another tank of 0.5 M3 with propane at 250 kPa, 4
    11·1 answer
  • Generally the primary source of all water supply is to be said as
    5·2 answers
  • Realize the function f(a, b, c, d, e) = Σ m(6, 7, 9, 11, 12, 13, 16, 17, 18, 20, 21, 23, 25, 28)using a 16-to-1 MUX with control
    13·1 answer
  • The toughness of a material does what, when it's been heated?​
    7·1 answer
  • Match the terms with the correct definitions.
    14·1 answer
  • . An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!