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Gala2k [10]
2 years ago
10

Explain why solar energy is considered an inexhaustible source of energy

Physics
1 answer:
Debora [2.8K]2 years ago
6 0

-- The amount of energy radiated by the sun is not affected at all by how much
of it we use or don't use.

-- The sun is not expected to quit shining on us, or to change its rate of energy
output, at any time within the next several many human lifetimes.

Solar energy is going to keep arriving here, continuously, at the same rate,
free of charge, for a long long time, whether we feel like using it or not.  It's
the closest thing to manna from heaven that has been seen around here for
thousands of years.


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What term describes the rate at which charge passes a point in a circuit
docker41 [41]
That's electric "current", usually described in 'Amperes'.

1 Ampere = 1 Coulomb per second ... "the rate at which
charge passes a point in the circuit".
8 0
3 years ago
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For the circuit shown in the figure(figure 1) find the current through each resistor. Express your answers using two significant
Angelina_Jolie [31]

The current flowing in each resistor of the circuit is 4 A.

<h3>Equivalent resistance of the series resistors</h3>

The equivalent resistance of the series circuit is calculated as follows;

6 Ω and 4 Ω are in series = 10 Ω

5 Ω and 10Ω are in series = 15 Ω

<h3>Effective resistance of the circuit</h3>

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\R = \frac{R_1R_2}{R_1 + R_2} \\\\R = \frac{10 \times 15}{10 + 15} \\\\R = 6 \ ohms

<h3>Current flowing in the circuit</h3>

V = IR

I = V/R

I = 24/6

I = 4 A

Learn more about resistors in parallel here: brainly.com/question/15121871

8 0
1 year ago
Increasing the two objects will cause the gravitational force between the objects to decrease.
ELEN [110]
This statement is false. Increasing the two objects' mass (I'm guessing) will actually increase their gravitational force. This is because of the equation:

F_g =  \frac{Gm_1m_2}{d^2}

If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.
4 0
3 years ago
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A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i
laiz [17]

(a) Let <em>x</em> be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work <em>W</em> done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to <em>x</em> is

<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)

(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so

<em>W</em> = ∆<em>K</em> = 0 - 1/2 <em>m</em> (0.835 m/s)²

Solve for <em>x</em> :

- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   <em>x</em> ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²

If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,

<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²

Solve for <em>v</em> :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²

==>   <em>v</em> ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(<em>k</em>/<em>m</em>) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0
2 years ago
1. Calculate the height of tree, 250 m away that produces
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Answer:

Explanation:

1.5/30 = x/250

x = 12.5 m

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