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Xelga [282]
3 years ago
11

Problem 2: Estimate the electric power requirement, in kW, of a 1,400 ft2floor area (three bedroomshome) with three occupants. U

sing your home power estimate, predict the power requirementsfor a city of 300,000 people. Use these results to estimate the area (inkm2)of silicon solar cells requiredto satisfy the community power requirements. Assume, thepower requirements for an average single family home of 3 are 108.4 x 106BTU per year and solar panels insolation
Physics
1 answer:
stira [4]3 years ago
8 0

Answer:

a) 3170 kw

b) 377 km^2

Explanation:

Estimate of electric power

a) Given :

Average power consumption for a family of 3 = 108.4 * 106 BTU per year  =  0.0317 kw = 31.7 watts

<u>The power requirement for a city of 300000 people </u>

= 31.7 watts * 100000 = 3170000 watts = 3170 kw

b) Given :

Average solar panel insulation = 8.4 W /m^2

<u>Estimate the area of silicon solar cells required to satisfy community power requirement</u>

= (1 * 3170) * (1000/8.4 )

= 377.380 * 10^3 m2 = 377 km^2

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Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont
koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

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7 0
2 years ago
6. Decelerating a plane at a uniform rate of -8 m/s2, a pilot stops the plane in 484 m. How
uysha [10]

Answer: 88 m/s

Explanation:

If we are talking about an acceleration at a uniform rate, we are dealing with constant acceleration, hence we can use the following equation:

{V_{f}}^{2}={V_{o}}^{2}+2ad (1)

Where:

V_{f}=0 Is the final velocity of the plane (we know it is zero because we are told the pilot stops the plane at a specific distance)

V_{o} Is the initial velocity of the plane

a=-8m/s^{2} is the constant acceleration of the plane

d=484m is the distance at which the plane stops

Isolating  V_{o} from (1):

V_{o}=\sqrt{-2ad} (2)

V_{o}=\sqrt{-2(-8m/s^{2})(484m)} (3)

Finally:

V_{o}=88m/s This is the veocity the plane had before braking began

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3 years ago
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from the NASA website, COPYRIGHT Jun 8, 2015

7 0
3 years ago
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