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BartSMP [9]
3 years ago
11

Wine and cheese are complement goods because they are consumed together. What would we expect to happen to the equilibrium quant

ity of cheese if the price of wine increased and all else is held constant?
Physics
1 answer:
notka56 [123]3 years ago
8 0

Explanation:

As wine and cheese are consumed together,it means that they both are complimentary goods.

As per law of demand,the fall in price of a complimentary good would increase the demand and shift the demand curve of the other to the right.

The equilibrium quantity of cheese would increase and shift to right when the price of wine falls.

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Two very long uniform lines of charge are parallel and are separated by 0.300 m. each line of charge has charge per unit length
Alenkasestr [34]

linear charge density of system of two line charges is given as

\lambda = 5.20 \muC/m

now as we know that electric field due to a line charge at some distance from it is given by

E = \frac{\lambda}{2\pi \epsilon_0 r}

so here we will first find the electric field of first line charge at the position of other line charge

E = \frac{5.20 * 10^{-6}}{2 \pi * 8.85 * 10^{-12}* 0.300}

E = 312000 N/C

now as we know that

F = qE

here q = charge on the line charge system at which force is required

E = electric field on that system of charge where force is required

now we can find the charge by

q = \lambda * L

q = 5.20 * 10^{-6}* 0.05 = 0.26 * 10^{-6} C

Now using the above formula

F = qE

F = 0.26 * 10^{-6} * 312000

F = 0.0811 N

so force on the part of wire is F = 0.0811 N

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3 years ago
Name three ways carbon and oxygen get into the environment.
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Humans,trees,plants   hope it helps

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4 years ago
A force of 200 N is required to keep an object sliding at a constant speed of 2 m/s across a rough floor. How much power is bein
WARRIOR [948]
Force x speed = power
200 x 2 = 400 watts
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3 years ago
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If a person can run 10 km in 2 hours, what is the person's speed? Include your work with your answer.
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A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 met
zlopas [31]

Answer:

b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

-kx - bv = ma

-kx -bdx/dt = md²x/dt².

Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

This is a second-order differential equation. The characteristic equation is thus,

D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}

For an overdamped system,

(b/m)^{2} - 4k/m} >   0

(b/m)^{2} >   4k/m}\\(b/m) >   \sqrt{4k/m}} \\(b/m) >   2\sqrt{k/m}} \\b >   2\sqrt{km}}

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s

6 0
3 years ago
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