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kaheart [24]
3 years ago
5

How is work related to potential energy and kinetic energy?

Physics
1 answer:
NemiM [27]3 years ago
3 0
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Featured snippet from the web
Work is the force on the object as it changes a distance. Interestingly, as work is done on an object, potential energy can be stored in that object. For example, if you carry a load up the stairs. Now that load will have potential energy that can be transformed into kinetic energy and so on
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Why is the electromagnetic spectrum called a spectrum?
Doss [256]

Answer & Explanation:

Scientists call them all electromagnetic radiation. The waves of energy are called electromagnetic (EM) because they have oscillating electric and magnetic fields. Scientists classify them by their frequency or wavelength, going from high to low frequency (short to long wavelength).

6 0
4 years ago
A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wh
DaniilM [7]
-- It takes the brick 8.9 seconds to reach the ground. 

-- At the instant of the "splat", it's falling at 89 m/s.

-- The mass doesn't matter. If not for air resistance, every object
    would fall at the same rate.  The answer is the same for a feather,
    a rubber chicken, a brick, or a school bus.
5 0
3 years ago
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as
Taya2010 [7]

Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

Change in internal energy

\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

- 5.02 * 101.3\\= 508.53Joules

q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

7 0
3 years ago
A weightlifter can lift 220lbs. (assume 1kg weighs 2.2lbs) If he lifts the weight 2m above the floor, how much work did he do?
Anton [14]
Convert 220 lb to kg.

220/2.2 = 100kg.

W = Fd (In this case, F is the weight)
W = (100)(2)
W = 200J

P = W/t
P = (200)/(1.2)
P = 166.67W
5 0
3 years ago
A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
4 years ago
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