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konstantin123 [22]
3 years ago
14

Elijah wants to make some changes to a game that he is creating. Elijah wants to move the bleachers to the right, the coaches to

the left, and the players to the center. What does Elijah want to change in the game?
the interface
the variable
the conditional loop
the editor

help soon pls-
Engineering
2 answers:
Ahat [919]3 years ago
7 0

Answer:

the editor

Explanation:

the variable is usually a number, and the conditional loop has nothign to do with movement/ where something is located

pshichka [43]3 years ago
3 0

Answer:

the editor

Explanation:

i got it right on the quiz!!!!!

hope this helps

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Describe with an example how corroded structures can lead to environment pollution? ​
raketka [301]
According to EonCoat, corrosion is the process of decay on a material caused by a chemical reaction with its environment. Corrosion of metal occurs when an exposed surface comes in contact with a gas or liquid, and the process is accelerated by exposure to warm temperature, acids, and salts.” (1)
Although the word ‘corrosion’ is used to describe the decay of metals, all natural and man-made materials are subject to decay, and the level of pollutants in the air can speed up this process.
5 0
3 years ago
About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

7 0
2 years ago
Describe ICP/OES in detail.
alex41 [277]

Answer:

ICP -OES stand for inductively coupled plasma optical emission spectroscopy

Explanation:

It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.

The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.

7 0
3 years ago
Read 2 more answers
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0
3 years ago
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