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Semmy [17]
3 years ago
7

What is control flow in programming?

Engineering
1 answer:
MakcuM [25]3 years ago
7 0

Answer:

The control flow is the order in which the computer executes statements in a script. Code is run in order from the first line in the file to the last line, unless the computer runs across the (extremely frequent) structures that change the control flow, such as conditionals and loops.

Explanation:

hope it helps.

brainliest pls

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A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above
creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

5 0
3 years ago
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How to cancel account
Anestetic [448]
You have to delete the app or delete your account
7 0
4 years ago
To combat overheating, keep a ___________ in your car.
Stells [14]

Answer:

Keep a gallon of water in your car.

Explanation:

Because.

4 0
3 years ago
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Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of
Orlov [11]

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

7 0
4 years ago
Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon
balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

torque =\frac{power}{2\pi N}

power=2\pi N\times T

P=2\times \pi \times2500 \times 4.5

P=70,685W

P=70.685 KW

power=\frac{work done}{time}

work done = power * time

                  = 70.685*30=2120.55J

                  = 2.12 kJ

7 0
4 years ago
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