Anonymous
Anonymous asked in Education & ReferenceHomework Help · 1 decade ago

physics help!!!!!!!!!!!!!!!?

alrighty can someone please help

me with physics. i honestly don't understand

it!

*what is the height of a 5.5 kg ball that has a 1715J of potential energy

*what is the mass of a man on top of a tower that is 43.2m talll if his potential energy is 50,803.2 Joules?

*how much force is required to move an object 12.5 m using oly 560J of work?

* What is the mass of a car moving 30m/s if it has a kinetic energy of 450,000J?

*if a man can work for 1 minute at a constant power rating of 22W, then how much work can that man do?

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  • 1 decade ago
    Favorite Answer

    1. 32 m or 31.8 m if preferred

    2. 120 kg or 119.9 kg if preferred

    3. 44.8 N

    4. 22 J

    1. For this one, I'll assume the ball is near Earth, not at some other place in the universe, and that the height is relative to the Earth's surface. Seems reasonable. At the Earth's surface, gravity has a strength of 9.81 m/sec^2, so if we put in the values we have:

    1715 J = (5.5 kg) * (x m) * (9.81 m/sec^2)

    That's because J = kg * m^2 / sec^2. Solve that for "x":

    1715 J = (5.5 kg) * (x) * (9.81 m/sec^2)....divide each side by 5.5 kg and by 9.81 m/sec^2

    (1715 J) / {(5.5 kg) * 9.81 m/sec^2)} = x....J = kg*m^2/sec^2

    (1715 kg*m^2/sec^2) / {(5.5 kg * 9.81 m/sec^2)}....rearrange numbers and the measurement units

    (1715 / (5.5 * 9.81) * (kg/kg * m^2/m * sec^2/sec^2)....do the arithmetic

    31.78574738 m

    which is 32 m taken to the proper 2 significant digits. (If you haven't done significant digits or don't use them unless specified, then use 31.8 m to convince your teacher you found the right answer.)

    2. This is very similar to 1. except you know the height and need the mass. So:

    50803.2 J = x * 43.2 m * 9.81 m/sec^2....divide each side by 43.2 m and by 9.81 m/sec^2

    x = (50803.2 J) / (43.2 m) * 9.81 m/sec^2)....J = kg*m^2/sec^2 and separate numbers and units

    x = {50803.2 / (43.2 * 9.81)} * {(kg*m^2/sec^2) / (m * m/sec^2)}....do the arithmetic

    x = 119.8776758 (kg * m^2/m^2 * sec^2/sec^2)

    x = 120 kg

    to the proper 3 significant digits, or 119.9 kg if you don't use significant digits.

    3. Joules equal Newtons * meters as well as the above which is mighty convenient as Newtons are a unit for force. So set up what you know like this:

    560 J = x * 12.5 m

    and solve for "x":

    560 J = x * 12.5 m....divide each side by 12.5 m

    x = (560 J) / (12.5 m)....substitute J = N * m

    x = (560 N * m) / (12.5 m)....separate numbers and units

    x = (560 / 12.5) * (N * m / m)....m/m = 1

    x = (560 / 12.5) N....do the arithmetic

    x = 44.8 N

    which is to the proper 3 significant digits.

    4. This time we can use the fact that 1 J = 1 watt-sec as we have a time and a power in the given information, conveniently already in watts (after all, minutes-to-seconds is fairly clear to most people so only the power units mattered to convenience, eh?). So this time, the work is "x":

    x = 1 min * 22 w....convert minutes to seconds with 1 = (60 sec) / (1 min)

    x = (1 min) * (60 sec / 1 min) * 22 w....rearrange numbers and units together

    x = (1 * 60 * 22 / 60) * (min * sec * w / min)....do the arithmetic

    x = 22 watt * sec * min/min....convert watt * sec using 1 = (1 J) / (1 watt * sec)

    x = 22 (watt * sec) * {1 J / (1 watt * sec)}....and cancelling units...

    x = 22 J * watt/watt * sec/sec....and so...

    x = 22 J

    which is to the proper 2 significant digits.

  • 4 years ago

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