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3 years ago

Solar energy is: expensive renewable inefficient during winter all of the above

Physics

2 answers:

creativ13 [48]3 years ago

3 0

Solar energy is free and renewable.

Its efficiency is determined by the photovoltaic or heat-exchanging equipment used, and the efficiency of the infrastructure built to interconnect them. None of that depends on the time of the year. For a given amount of solar energy absorbed by the collector, the amount of useful output energy doesn't change with the season.

Mazyrski [523]3 years ago

3 0

All of the above is the right answer for Odesseyware.

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1. A student demonstrates electromagnetic induction using a

Katen [24]

answer is :D it would be a great answer

4 0

3 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

3 years ago

Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, $\theta=25^{\circ}$

Time, t = 3 s

Let h is the height of the hill from the horizontal,

$h=l\ sin\theta$

$h=12\times \ sin(25)$

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

$P=\dfrac{E}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{55.8\times 9.8\times 5.07}{3}$

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

3 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 8.84 μC. A conducting spherical shell of inner radius 4.00 cm and ou

AlladinOne [14]

Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5

Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.

Region 1; 0<r<2

∫E.ds=Qinside the gaussian surface/ε0

inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.

Region 2; 2<r<4;

E.4*π*r^2=8,84/ε0

E=8,84/(4*π*ε0*r^2)

Region 3; 4<r<5

E=0 because is inside the conductor.

Finally

Region 4; r>5

E.4*π*r^2=(8,84-2.02)/ε0

6 0

3 years ago

A car travels in a flat circle of radius R. At a certain instant the velocity of the car is 24 m/s west and the total accelerati

lisov135 [29]

Answer with Explanation:

We are given that

Velocity,v=24 m/s west

Acceleration,a= $2.5m/s^2$

$\theta=53^{\circ}$ (N of W)

Horizontal component of acceleration=Tangential acceleration

Tangential acceleration, $a_x=acos\theta=2.5cos53=1.505 m/s^2$

Radial acceleration=Vertical acceleration= $a_y=asin\theta$

$a_y=2.5sin53=1.998 m/s^2$

Radial acceleration, $a_y=\frac{v^2}{R}$

$1.998=\frac{(24)^2}{R}$

$R=\frac{(24)^2}{1.998}$

$R=288.29 m$

Time, $t=\frac{2\pi R}{velocity}$

$t=\frac{2\pi(288.29)}{24}$

$t=75.47 s$

8 0

3 years ago