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3 years ago

The principal limitation of wind power is unpredictability. Please select the best answer from the choices provided

Physics

2 answers:

user100 [1]3 years ago

7 0

Answer:

A. True

Explanation:

Lady_Fox [76]3 years ago

6 0

<span>The correct answer is A: True. Although the ability to forecast weather has increased over the years, the ability to predict when and how much wind will occur at any given time and place is difficult. Weather conditions are subject to constant change and while some areas are known to have more wind than others, there is no guarantee about how much wind there will be as weather patterns vary in consistency.</span>

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mylen [45]

The experiments will involve two billiard balls of known masses, m₁ and m₂, and velocities u₁ and u₂. The two are allowed to collide and the velocities of the balls after the collision v₁ and v₂ are recorded.

The momentum before and after the collision is then calculated as follows:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

<h3>What is the statement of the law of conservation of momentum?</h3>

The law of the conservation of momentum states that the momentum before and after collision in a system of colliding bodies is conserved

The momentum of a body is calculated using the formula below:

Momentum = mass * velocity.

Hence, for the two billiard balls, the momentum before and after the collision is conserved.

Learn more about momentum at: brainly.com/question/1042017

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3 0

1 year ago

6. And what's up with ChiN?

n200080 [17]

I don’t know what’s up with chin man ??

4 0

2 years ago

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

Marizza181 [45]

The energy stored in the membrane is $6.44\cdot 10^{-14} J$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

$A=4.50\cdot 10^{-9} m^2$

$d=8.1\cdot 10^{-9} m$

Substituting, we find its capacitance:

$C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F$

Now we can find the energy stored: for a capacitor, it is given by

$U=\frac{1}{2}CV^2$

where

$C=2.26\cdot 10^{-11} F$ is the capacitance

$V=7.55\cdot 10^{-2} V$ is the potential difference

Substituting,

$U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J$

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

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6 0

3 years ago

What is the critical angle of light traveling from benzene (n=1.501) into air?

Taya2010 [7]

The critical angle formula should be: sin^-1(1/n)

where "n" is 1.501 into the air

<span>The critical angle of light travelling from benzene, happens because the larger angles of incidence from the inside of the benzene has experienced the total internal reflection. </span>

4 0

3 years ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago