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love history [14]
3 years ago
7

Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more

gravitational potential energy would you gain if you took the longer path
Physics
1 answer:
umka2103 [35]3 years ago
6 0

Answer:

None.

Explanation:

  • Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
  • As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

       \Delta U = U_{f} -U_{0} = m*g*h - 0  = m*g*h

  • As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.
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Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible
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Inertial I = mr^2 where r = distance from axis of rotation, while m is the mass of the object.

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Which statement about dwarf planet locations is correct?
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Q 24, 25, 26 i dont get them and need answers for it
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Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

DU = Q-W\\

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]  

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

Q = W\\

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

E_{p}=Q\\\\E_{p}=m*g*h

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]

The heat developed can be calculated by means of the following equation.

Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

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