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Leona [35]
3 years ago
8

A student pushed a 100 N bicycle over a distance of 15 m in 5 s. calculate the power generated.

Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

The catch in this one is:  We don't know how much <u>force</u> the student used to push the bike.  

It wasn't necessarily the 100N.  That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium.  The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push.  That's what's so great about wheels !  That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel".  I looked at it, I turned it over and I looked on all sides. I thought it was too simple.  I didn't know then it was elegant. I threw him out.  I was so dumb.  I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ?  Stay tuned:

-- The bike covered 15 meters in 5 seconds.  Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.  

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

<em>Power = 36.73 watts</em>

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Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

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Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

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Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

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Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

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3 years ago
Amotor supplied by 240V requires 12A to lift a 2000 lb at a rate of 25 ft/min, me power input to the motor is-
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Answer:

Power input, P = 2880 watts

Explanation:

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So, the power input of the motor is 2880 watts. Hence, this is the required solution.

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At a certain instant, coil A is in a 10-T external magnetic field and coil B is in a 1-T external magnetic field. Both coils hav
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Answer:

A) coil A

Explanation:

According to Faraday, Induced emf is given as;

E.M.F = ΔФ/t

ΔФ = BACosθ

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θ is the angle between the magnetic field, B, and the normal to the loop of area A

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From the equation above, induced emf depends on the strength of the magnetic field.

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anzhelika [568]

Answer:

10°C

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To convert °F to °C, we use the formula:

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So, to convert 50°F to the equivalent  in °C, we can proceed as follows:  

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°C = 9.999999999... ≈ 10 (5/9 ≈0.555555...)

So, 50°F is equivalent to 10°C.

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