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3 years ago

Which characteristics is common to renewable energy resources?

1 answer:

4 0

Renewable Energy resources can easily be produced as soon as they are used.

It is sufficient in supply and it is produced easily within a relatively short time.

Example are: Solar Energy , Wind Energy, Hydro Energy.

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One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

According to the what the universe's total amount of energy does not change

RideAnS [48]

The Law of Conservation of Energy

8 0

3 years ago

Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you

gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

$\mu_{0} = 4\pi * 10^{-7}$

The inductance of a solenoid is given by:

$L = \frac{\mu_{0}N^{2} A }{l}$

$500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns$

8 0

2 years ago

2) A car travels 5 miles north, and then 10 miles south. What is the person's DISTANCE and

Liono4ka [1.6K]

Answer:

distance traveled is 15 mi

displacement is 5 mi

Explanation:

Distance takes time into account and adds up all the tiny displacements during the entire period of the trip.

Displacement ignores time and looks only at the change in position from the starting point to the ending point.

6 0

2 years ago

A pelican flying along a horizontal path drops

Ludmilka [50]

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s$

The initial horizontal speed of the pelican is given by;

X = vₓt

vₓ = X / t

vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

8 0

3 years ago