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3 years ago

What is the real name of the town that Bud & Bugs stayed in while they were waiting for the train?

Engineering

1 answer:

Leona [35]3 years ago

6 0

Answer: Hooverville

Explanation:

Bud, not Buddy is a book about a ten years old boy who ran away from home where he stayed with his foster parents. He was treated unfairly and beaten.

He left home to seek a better living and hope that he will find his father as well. He met another orphan named Bugs and they decided to go to Hooverville so that they can get a train going to California.

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Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s

ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0

3 years ago

What are some common ways of converting fuel to useful energy? Check all that apply.

Rainbow [258]

Answer: Burning it.

Explanation: When your car moves its due to the burning of fuel, works like a turbine the molecules evaporate and turn a turbine like object.

3 0

3 years ago

Read 2 more answers

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Direction: List down or enumerate the type of outlets you want to install on your dream house. Also indicate the quantity (in pi

Lorico [155]

Answer:

Apartment outlet 4pcs

Explanation:

6 0

3 years ago

A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is

Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

6 0

3 years ago