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Greeley [361]
2 years ago
11

B. Is the “Loading Time” of any online application a functional or a non-functional requirement? Can the requirement engineers s

pecify this property before the system is actually implemented, how?
Engineering
1 answer:
Oksi-84 [34.3K]2 years ago
4 0

Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

Explanation:

non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

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stich3 [128]
Can you be a bit more specific plz and that will let me identify the answer
6 0
3 years ago
Which of these cars traveled faster during time interval <br> please show solution
Mazyrski [523]

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

3 0
3 years ago
Which starting circuit uses fuses, switches, and smaller wires to energize a relay and solenoid?
zysi [14]

Answer:

Option B (Starter Control Circuit) is the right option.

Explanation:

  • This same switching is normally put upon this isolated side of something like the transmission Arduino microcontroller throughout the configuration that is using the ignition just to command the broadcast.
  • It uses a secondary relay isolated to regulate electrical current throughout the solenoid starting system.

All other given options are not related to the given instance. So the above option is correct.

4 0
3 years ago
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

\Delta S=\dfrac{Q}{T}

Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0
3 years ago
The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both
DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

                               L^2 = x_a^2 + y_b^2   ....... 1

                               y_b = sqrt( 20^2 - 16^2 )

                               y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

                               0 = 2*x_a*v_a + 2*y_b*v_b

                               0 = x_a*v_a + y_b*v_b   ..... 2

                               0 = 16*36 + 12*v_b

                               v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

                               0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

                               0 = 9 + 4*a_a/3 + 16 + a_b

                               4*a_a/3 + a_b = -25

                               4*a_a + 3*a_b = -75  .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

                               F_x = m_a*a_a

                               P - R_r*16/20 = m_a*a_a

                               

- For Slider B, Apply Newton's second law of motion in y direction:

                               F_y = m_b*a_b

                               - R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

                               P - 4*m_b*a_b / 3 = m_a*a_a

                               3P = 3*m_a*a_a + 4*m_b*a_b  ... 4

- Where,                  P = Is the force acting on slider A

                               P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.                    

5 0
2 years ago
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