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3 years ago

What are the limitations of using solar panels?

1 answer:

8 0

Location & Sunlight Availability.
Solar Panels use a large amount of space.
The Sun isn't always present.
Solar Energy is Inefficient.
There is an overlooked Pollution & Environmental Impact.
Expensive Energy Storage.
High Initial Cost.

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I want to take a picture of the question and get the answer

OleMash [197]

When you take a picture it pulls up people who asked the same question and they’ll have a answer you can use

5 0

3 years ago

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$

Substitute the given values and solve the speed;

$v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s$

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

7 0

3 years ago

To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

Karolina [17]

Answer:

a) T² = ( $\frac{4\pi ^2}{GM}$ ) r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

F = ma

in this case the acceleration is centripetal

a = v² / r

the linear and angular variable are related

v = w r

we substitute

a = w² r

force is the universal force of attraction

F = $G \frac{m M}{r^2}$

we substitute

$G \frac{m M}{r^2} = m w^2 r$

w² = $\frac{GM}{r^3}$

angular velocity is related to frequency and period

w = 2π f = 2π / T

we substitute

$( \frac{2\pi }{T} ) = \frac{GM}{r^3}$

the final equation is

T² = () r³

b) the speed of the orbit can be found

v = w r

v = $\sqrt{\frac{GM}{r^3} } \ r$

v = $\sqrt{\frac{GM}{r} }$

in this case the dependency is the inverse of the root of the distance

Kinetic energy

K = ½ M v²

K = ½ M GM / r

K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

U =

U = -G mM / r

dependency is the inverse of distance

Angular momentum

L = r x p

for a circular orbit

L = r p = r Mv

L =

The angular momentum depends directly on the root of the distance

8 0

3 years ago

What is the simplest way to build a graph of kinetic and potential energy?

Elena-2011 [213]

Answer:

analize the levels of kinetic and potential energy and look up a guide to graph it and follow that

Explanation:

5 0

3 years ago

If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?

astraxan [27]

Mass of the object is given as

$m = 1417 g = 1.417 kg$

now the speed of object is given as

$v = \frac{d}{t}$

here we know that

$d = 47 m$

$t = 90 s$

now we will have

$v = \frac{47}{90} = 0.52 m/s$

now we will have kinetic energy of the object as

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(1.417)(0.52)^2$

$KE = 0.19 J$

now the power is defined as rate of energy

so here we can find power as

$P = \frac{KE}{t}$

$P = \frac{0.19}{90} = 2.14\times 10^{-3} W$

so above is the power used for the object

3 0

3 years ago

What are the limitations of using solar panels?​

What are the limitations of using solar panels?