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Gwar [14]
2 years ago
14

Request for proposal (RFP) is a type of document that contains the information and proposals mostly through the bidding process.

This document is regarding the valuable assets, services, entity, commodity, etc.
Engineering
1 answer:
Art [367]2 years ago
8 0

Answer:

Answer to the following is as follows;

Explanation:

A request for proposal is a documentation that invites prospective contractors to submit business opportunities to an agency or corporation interested in procuring a commodities, product, or valuable resource through a bid procedure.

A request for proposal (RFP) is a commercial document that introduces a project, defines it, and invites eligible contractors to compete on its completion.

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Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
The current in a 20 mH inductor is known to be: 푖푖=40푚푚푚푚푡푡≤0푖푖=푚푚1푒푒−10,000푡푡+푚푚2푒푒−40,000푡푡푚푚푡푡≥0The voltage across the induct
Anni [7]

Answer:

a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.

Explanation:

The question text is corrupted. I found the complete question on the web and it goes as follow:

The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.

a. Find the numerical expressions for i and v for t>0.

b. Specify the time intervals when the inductor is storing energy and is delivering energy.

A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:

i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))

40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)

40*10^(-3) = (A1)*1 + (A2)*1

40*10^(-3) = A1 + A2

A1 + A2 = 40*10^(-3)

Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:

V = L*di/dt

We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:

di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt

di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)

di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)

By applying t = 0 to this expression we have:

di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)

di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0

di/dt (at t = 0) = -10,000*A1- 40,000*A2

We can now use the voltage equation for the inductor at t=0, that is:

v = L di/dt (at t=0)

68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)

68 = -400*A1 -800*A2

-400*A1 - 800*A2 = 68

We now have a system with two equations and two variable, therefore we can solve it for both:

A1 + A2 = 40*10^(-3)

-400*A1 - 800*A2 = 68

Using the first equation we have:

A1 = 40*10^(-3) - A2

We can apply this to the second equation to solve for A2:

-400*[40*10^(-3) - A2] - 800*A2 = 68

-1.6 + 400*A2 - 800*A2 = 68

-1.6 -400*A2 = 68

-400*A2 = 68 + 1.6

A2 = 69.6/400 = 0.174

We use this value of A2 to calculate A1:

A1 = 40*10^(-3) - 0.174 = -0.134

Applying these values on the expression we have the equations for both the current and tension on the inductor:

i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]

v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]

v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.

8 0
3 years ago
A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1
Gelneren [198K]

Answer:

final temperature is 424.8 K

so correct option is e 424.8 K

Explanation:

given data

pressure p1 = 1 bar

pressure p2 = 5 bar

index k = 1.3

temperature t1 = 20°C = 293 k

to find out

final temperature  t2

solution

we have given compression is reversible and has an index k

so we can say temperature is

\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }  ...........1

put here all these value and we get t2

\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }

t2 = 424.8

final temperature is 424.8 K

so correct option is e

5 0
3 years ago
Which allows a user to run applications on a computing device? Group of answer choices Application software CSS Operating system
sveticcg [70]

Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user.  The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

Cheers.

4 0
3 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
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