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dmitriy555 [2]
3 years ago
15

What are two ways in which the suns energy can be captured and used?

Physics
2 answers:
Nezavi [6.7K]3 years ago
8 0
It can be used by giving you energy and buring calories
My name is Ann [436]3 years ago
7 0

The oldest way ... the way we've been using as long as we've been
walking on the Earth ... has been to use plants.  Plants sit out in the
sun all day, capturing its energy and using it to make chemical compounds. 
Then we come along, cut the plants down, and eat them.  Our bodies
rip the chemical compounds apart and suck the solar energy out of them,
and then we use the energy to walk around, sing, and play video games.  

Another way to capture the sun's energy is to build a dam across a creek
or a river, so that the water can't flow past it.  You see, it was the sun's
energy that evaporated the water from the ocean and lifted it high into
the sky, giving it a lot of potential energy.  The rain falls on high ground,
up in the mountains, so the water still has most of that potential energy
as it drizzles down the river to the ocean.  If we catch it on its way, we
can use some of that potential energy to turn wheels, grind our grain,
turn our hydroelectric turbines to get electrical energy ... all kinds of jobs. 

A modern, recent new way to capture some of the sun's energy is to use
photovoltaic cells.  Those are the flat blue things that you see on roofs
everywhere.  When the sun shines on them, they convert some of its
energy into electrical energy.  We use some of what they produce, and
we store the rest in giant batteries, to use when the sun is not there.
 
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Work is a transfer of (1 point) energy. force. mass. motion.
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Work is the transfer of energy.
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3 years ago
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Where do you feel that you are traveling at the fastest speed when on the swing?
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Answer:

C

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

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3 years ago
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Which is an example of gaining a static charge by conduction?
lianna [129]

Answer:

shuffling shoes

Explanation:

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3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
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Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci
diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)×\frac{\pi}{180}/3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

r₁ = 20 in = 1.67 ft

= (1.667 - 1.195)/0.47

0.472/0.47

u= 1.00ft/s

acceleration at collar p

a=rω²

= 1.67 × 3.65²

a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

= 22.5j + 0 + 7.3i

√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

a= 23.65 ft/s²

4 0
3 years ago
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