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1 year ago

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The price of coffee fell sharply last month, while the quantity sold remained the same. Five people suggest various explanations

: Antonio: Demand decreased, but it was perfectly inelastic. Caroline: Demand decreased, but supply was perfectly inelastic. Dmitri: Demand decreased, but supply increased at the same time. Frances: Supply increased, but demand was perfectly inelastic. Jake: Supply increased, but demand was unit elastic. Who could possibly be right? Check all that apply.
Rajiv
Simone
Yakov
Ana
Charles

1 answer:

miss Akunina [59]1 year ago

7 0

Answer:

France. True Supply increased and demand did not adapt

Explanation:

If the price of coffee decreases it may be due to two causes

- Increase of the offer

- decrease in demand

- both simultaneously.

Let's examine the statement. They tell us that the amount sold remained constant, so there is no decrease in demand, therefore the only cause may be the increase in supply.

Let's check the answers

Antonio False. The demand is constant.

Carolina. False. The demand is constant.

Dmitri False. The demand is constant.

France. True Supply increased and demand did not adapt

Jake. False. Demand did not adapt to the increase in supply

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Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E

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Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

$H = \vec d$

work done is given as

$W = \vec F . \vec d$

Potential energy is given as

$PE_g = mgh$

force due to gravity is given as

$\vec F_g = mg$

now here if we plug in the value of distance and force in the formula of work done then we will have

$W = (mg)(h)$

so here we got

$W = PE_g$

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

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In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

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W = P_p x Q

W = 635539.32 x 0.015

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Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as stron

WARRIOR [948]

Answer:

The speed of ejection is $2.06\times 10^{4}\ m/s$

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, $\rho = 3\times 10^{4}\ kg/m^{3}$

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

$U_{B} = \frac{B^{2}}{2\mu_{o}}$

This energy density equals the kinetic energy supplied by the field.

Thus

$KE = U_{B}$

$\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}$

where

m = mass of the sunspot in $1\ m^{3}$ = $3\times 10^{- 4}\ kg/m^{3}$

$v = \frac{B}{\sqrt{\mu_{o}m}}$

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matrenka [14]

Answer:

A) F = - 8.5 10² N, B) I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

I = Δp

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v₀ = 8.50 m / s

v_f = -8.50 m / s

F t = m v_f -m v₀

F = $m \frac{(v_f - v_o)}{t}$

let's calculate

F = $1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}$

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v² = v₀² - 2g (y- y₀)

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v = $\sqrt{2g y_o}$

calculate

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I = Δp

I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

I = 1.50 (0-14)

I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

I = 21 N s

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zzz [600]

Answer:

Ordinal

Explanation:

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The other forms of data include the nominal which simply qualifies the data, the interval which qualifies the data but which the differences between the data can be obtained, and of course the data has no starting point. The ratio scale which is similar to the interval scale but which the ratios between the data obtained can be compared.

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