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3 years ago

What are the limitations of using solar panels?

Physics

1 answer:

Anit [1.1K]3 years ago

7 0

Answer:

The amount of the sun energy that could be collected.

Explanation:

Some limitations are, the amount of the sun's energy that could be collected the radiation of sun is nearly fixed. The place where we can put the solar panels are also limited.

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9. The acceleration due to gravity on the

laiz [17]

W=mg
W=75(1.6)
W=120N
a. 120N

6 0

3 years ago

If a ball is rolling across a floor at a constant speed is it accelerating

KATRIN_1 [288]

You already told us that its speed is constant. That's one part of acceleration.

The other part of acceleration is the direction it's moving.

If it's rolling in a straight line, then there's no acceleration.

If it's curving left or right, then that's acceleration.

8 0

3 years ago

Find the intensity of electromagnetic radiation at the surface of the sun (radius r=R=6.96×105kmr=R=6.96×105km). Ignore any scat

alisha [4.7K]

Answer:

$I = 4.46*10^{16}W/m^2$ .

Explanation:

Intensity $I$ of the electromagnetic radiation is given by

$I = \dfrac{P}{4\pi r^2},$

where $r$ is the distance from the EM source (the center of the sun, in our case), and $P$ is the power output of the sun and it has the value

$P = 3.9 *10^{26}W.$

Since the radius of the sun in meters is $r = 6.96*10^8km$ , the intensity $I$ of the electromagnetic radiation at the surface of the sun is

$I = \dfrac{3.9*10^{26}W}{4\pi (6.96*10^8m)^2}\\\\\boxed{ I = 4.46*10^{16}W/m^2}$

The intensity of the electromagnetic radiation at the surface of the sun is $I = 4.46*10^{16}W/m^2$ .

7 0

3 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous

Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

$v=\frac{2\pi \times r}{24\times 3600}$

The Centripetal force of the satellite is balanced by the universal gravitational force

$m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m$

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0

2 years ago