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Natasha2012 [34]
3 years ago
14

A typical person can maintain a steady energy expenditure of 400 W on a bicycle. Assuming a typical efficiency for the body and

a generator that is 85% efficient, what useful electric power could you produce with a bicycle-powered generator? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

we will get 340 W of output power when there is an input energy of 400 W with 85% efficient generator

Explanation:

As we know that efficiency of the generator is given as 85 %

so we can define efficiency of the generator as

efficiency = \frac{work \: output}{input\: energy} \times 100

so here we know that efficiency is 85 %

so it shows that it will given us 85% of output work when we give it some energy or input

So we will have

85 = \frac{W}{400}\times 100

W = \frac{85 \times 400}{100}

W = 340 W

So we will get 340 W of output power when there is an input energy of 400 W with 85% efficient generator

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How do I calculate the value of the slope graph​
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Slope is calculated using the equation: slope= rise/run (y/x). Take take y value of a point and put it over the x value of the same point and then simplify the fraction.

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3 years ago
A student launches a small 0.5 kg rocket with an initial speed of 30 m/s at an angle of 60°. Approximately how much time will th
Elena L [17]

Answer:

t=5.30s

Explanation:

The high reached by a proyectile in an uniformly accelerated motion is given by:

y=v_{0y}t-\frac{gt^2}{2}

The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that v_{0y}=v_0sin\theta. Solving for t:

0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s

5 0
4 years ago
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

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4 years ago
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Answer:

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