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3 years ago

25 points and brainliest is it A, B, C, D

Engineering

2 answers:

Bogdan [553]3 years ago

4 0

Answer:

SIR IT IS D HOPE THIS HELPS (☞ﾟヮﾟ)☞☜(ﾟヮﾟ☜)

Explanation:

Shkiper50 [21]3 years ago

4 0

I think B because if he is mixing chemicals he should have the goggles on his eyes not around his neck

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AC motor characteristics require the applied voltage to be proportionally adjusted by an AC drive whenever the frequency is chan

Margarita [4]

The answer is false

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3 years ago

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A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro

Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= $\frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt$

$= \int\limits^a_b {(5-3t)} \, dt$

$5t - \frac{3t^2}{2} +c$

v2 = 5x2 - 3x2 + c

= 10-6+c

= 4+c

$s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c$

S2 - S1

$=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)$

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

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2 years ago

Which of these is the coarsest grit abrasive that may be used on aluminum?

natali 33 [55]

Answer:

80grit

Explanation:

80 grit is coarsest grit that may be used on aluminum

The lowest grit sizes range from 40 to 60. From the given options 80 grit is practically available grit.

What is a sandpaper used for?

They are essentially used for surface preparation. Sandpaper is produced in a range of grit sizes and is used to remove material from surfaces, either to make them smoother (for example, in painting and wood finishing), to remove a layer of material (such as old paint), or sometimes to make the surface rougher (for example, as a preparation for gluing).

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3 years ago

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Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,

aalyn [17]

Answer:

import java.util.Scanner;

public class InputExample {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int birthMonth;

int birthYear;

birthMonth = scnr.nextInt();

birthYear = scnr.nextInt();

System.out.println(birthMonth+"/"+birthYear);

}

3 0

3 years ago

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

3 years ago