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4 years ago

Select all that apply. Select the items that describe an economy at equilibrium.

Physics

1 answer:

ella [17]4 years ago

8 0

The correct answers are:

B) Producers supply enough goods for consumers;

D) Supply equals demand;

In the economics, the equilibrium state basically means that the supply in the market is well balanced with the demand on the market, thus leading to an ideal economic state within the economy. The equilibrium will essentially mean that the supply is equal to the demand, thus there will be maximum efficiency. Also, since the producers are managing to supply just the right amount of goods to the consumers, everyone will benefit, with the consumers always have what they want, and the producers always have all of their products sold.

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A small Styrofoam bead with a charge of −60.0 nC is at the center of an insulating plastic spherical shell with an inner radius

Romashka-Z-Leto [24]

Answer:

speed of the proton is 6.286 × $10^{5}$ m/s

Explanation:

given data

charge q= −60.0 nC

inner radius a = 20.0 cm

outer radius b = 24.0 cm

charge density ρ = −2.05 µC/m³

to find out

What is the speed of the proton

solution

we know that force on the proton due to this electric field is express as

F = q × E ...................1

here F is force and q is charge and E is electric filed so

if v be the speed of the proton in circular orbit than force will be

F = $\frac{mv^2}{b}$ ....................2

from equation 1 and 2

q × E = $\frac{mv^2}{b}$ .......................3

here total charge Q on shell is

Q = ρ × V

here ρ is density and V is volume

Q = $\rho * \frac{4}{3} \pi (b^3-a^3)$

put here value

Q = $-2.05*10^{-6} \frac{4}{3} \pi (0.24^3-0.20^3)$

Q = 50.01 × $10^{-9}$ C

and

total charge enclosed by Gaussian surface is

qin = q + Q

qin = −60 × $10^{-9}$ C - 50.01 × $10^{-9}$ C

qin = - 110.01 × $10^{-9}$ C

and

from Gauss law

$\oint E*A = \frac{\left | qin \right | }{\epsilon }$

E ×4×π×b² = $\frac{110.01*10^{-9} }{\epsilon }$

E = $\frac{110.01*10^{-9} }{\epsilon *4 *\pi *b^2[tex]}$

E = $\frac{9*10^9 * 110.01*10^{-9} }{0.24^2}$

E = 17189.06 N/C

from equation 3

q × E = $\frac{mv^2}{b}$

v = $\sqrt{\frac{q*E*b}{m}}$

v = $\sqrt{\frac{1.6*10^{-19}*17189.06*0.24}{1.67*10^{-27}}}$

v = 6.286 × $10^{5}$ m/s

so speed of the proton is 6.286 × $10^{5}$ m/s

4 0

3 years ago

A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0

3 years ago

If white light falls on a piece of white paper in a dark room, the paper appears white to the eye. This is because the paper ref

Sindrei [870]

It would appear Red because as stated above, the paper reflects all the colors of light that fall on it.

6 0

3 years ago

Read 2 more answers

8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro

omeli [17]

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

$X = \frac{MsXs+McXc}{Mw+Mc} \\\\$

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

$X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\$

The center of gravity lies at the center of this boat. Therefore,

$Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5$

5.00 - 1. 00 = 4 meters

$\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\$

To get the distance that is moved by this canoe

distance = 3.143-1.857

= 1.286

≈ 1.29 meters

The distance that the canoe moves in this process is 1.29 meters.

6 0

2 years ago

Consider an airplane with a wing area of 240 ft2 , a span of 44 ft, an Oswald efficiency of 0.75, and a zero-lift drag coefficie

Darya [45]

Answer:

Drag force=1/2Cd*Area*density*V^2

A=240 ft^2

cd=0.75

Drage force=675 lbs

6 0

4 years ago