All categories

3 years ago

50 points

Engineering

1 answer:

Burka [1]3 years ago

4 0

Answer:

Water vapor

Explanation:

When water is in a vapor it tends to rise to a higher point. Because of this it would be able to reach the top of a building.

You might be interested in

PLEASE HELP ME! THIS QUESTION IS PART OF AN ASSIGNMENT THAT IS DUE TODAY! I DON'T UNDERSTAND HOW TO DO THIS QUESTION! PLEASE HEL

matrenka [14]

Magnitudes of the currents are i1= 0.00104A , i2= 0.003641A , i3= 0.00508A.

Explanation:

Using superposition theorem,

remove the E1 voltage supply source and calculate resistance across it.

From the circuit given, as the resistances R1 and R2 are parallel

using the formula R1//R2=(R1.R2/(R1+R3)

V1= ((r1||r2)/(r1+r2||r3))*E1

v1 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*10v

v1= 2.3v

v2 = ((r1||r2)/(r1+r2||r3))*E2

v2 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*5v

v2= 1.161v

V = V1+V2

=> 2.3 + 1.161

=> 3.461v

Magnitudes of the currents can be found by i=v/r

i1 = v/r1

=> 3.461/3.3kΩ

=>0.00104A

i2=2.89/1kΩ

=>0.003461A

i3=2.89/680Ω

=> 0.00508A.

Therefore the magnitudes of the currents are i1, i2, and i3.

7 0

3 years ago

Write a program. Submit a file named height.c Amusement parks have minimum heights for their rides. Create a program that adds u

Natasha_Volkova [10]

Answer:

you ok

Explanation:

try it

8 0

3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

PDAs with two stacks are strictly more powerful than PDAs with one stack. Prove that 2-stack PDAs are not a valid model for CFLs

tangare [24]

Answer:

?pooooooooooooooooooooooop

Explanation:

?pooooooooooooooooooooooop

6 0

3 years ago

Match each context to the type of the law that is most suitable for it.

Bas_tet [7]

Answer:

sorry i dont understand the answer

Explanation:

but i think its a xd jk psml lol

5 0

3 years ago