Kinetic energy into electrical energy

Physics

2 answers:

sleet_krkn [62]4 years ago

7 0

An example of that would be with wind turbines, but I see no question here to answer.

jeyben [28]4 years ago

3 0

A wind turbine is a machine for converting the kinetic energy in wind into electrical energy. Moving air (wind) is used to rotate a machine called a turbine. Inside the wind turbine is an electrical generator. It's the electricalgenerator that transforms the kinetic energy into electrical energy.

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Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Ear

vagabundo [1.1K]

Answer:

The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. RTT = 2.56s

Explanation:

Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m

Speed of data travel = speed of light ≈ 3 x 10⁸m/s

As, v=d/t

t=d/v

$t=\frac{3.85*10^{8} }{3*10^{8}}$

t=1.28s

RTT = Double of single way time taken = 2x1.28

RTT=2.56s

7 0

3 years ago

a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .