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3 years ago

I need help coding a game what do i do first?

Engineering

1 answer:

Rufina [12.5K]3 years ago

8 0

Answer:

more information please

Explanation:

please

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When an electron in a valence band is raised to a conduction band by sufficient light energy, semiconductors start conducting __

garri49 [273]

Answer:

This band gap also allows semiconductors to convert light into electricity in photovoltaic cells and to emit light as LEDs when made into certain types of diodes. Both these processes rely on the energy absorbed or released by electrons moving between the conduction and valence bands.

Explanation:

On the internet

4 0

3 years ago

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

padilas [110]

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

6 0

3 years ago

Read 2 more answers

Consider 2 hosts, Host A and Host B, transmitting a large file to a Server C over a bottleneck link with a rate of R kbps.

nevsk [136]

Answer:

i want coins sorry use a calculator or sum

Explanation:

5 0

3 years ago

Two parallel Rivers (A and B) are separated by confined and unconfined aquifer estimate the RATE of seepage of river A to River

gizmo_the_mogwai [7]

Answer:

Explanation:

A confined aquifer is one that's impermeable and does not allow seepage's of water through it.

An unconfined aquifer is a body of water that has permeable membranes which allows passage of water through it.

There will be little or no transmission of water trough these bodies of water.

3 0

3 years ago

A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s

monitta

Answer:

a. $\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$ which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

$v_1 = \dfrac{4}{3} \times \pi \times r^3$

$\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3$

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

$\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}$

We have;

$\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$

$\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )$

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

$\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}$