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oee [108]
3 years ago
5

The superheroine Xanaxa, who has a mass of 67.5 kg , is pursuing the 70.3 kg archvillain Lexlax. She leaps from the ground to th

e top of a 179 m high building then dives off it and comes to rest at the bottom of a 15.9 m deep excavation where she finds Lexlax and neutralizes him. Does all this bring about a net gain or a net loss of gravitational potential energy?
Physics
1 answer:
Paladinen [302]3 years ago
4 0

Answer:

there is a net loss of potential energy  is Ut = -0.11 10⁵ J

Explanation:

The gravitational potential energy is

        ΔU = mg (y2- y1)

Where y1 is the initial position where the energy is defined as zero, in general in problems on the Earth's surface the reference system is taken at the lowest point, whereby y1 = 0 and consequently U1 = 0, this It is a matter of comfort since what matters are the changes in energy and not its value at one point explained.

Let's apply these concepts to our problem.

Jump from the ground to 179 m, let's calculate the energy change

     ΔU1 = mg (y2-y1)

     ΔU1 = 67.5 9.8 (179 -0)

     ΔU1 = 1.18 10⁵ J           gains energy

Dives up to Y3 = -15.9m below ground, therefore, this height is negative

The total height is y = 179 - (-15.9) = 194.9 m

       ΔU2 = mg (y3-y2)

These are the points where the dive begins and where it arrives

       ΔDU2 = 67.5 9.8 (- 194.9)

       ΔU2 = -1.29 105 J        loses energy

The total change in potential energy is

       Ut = ΔU1 + ΔU2

       Ut = 1.18 10⁵ - 1.29 10⁵

       Ut = -0.11 10⁵ J

Therefore, there is a net loss of potential energy

You can also calculate ΔU = mg (Δy)

Where ΔY is the change in body position, that is -15.9 m

         ΔU = 67.5 9.8 (-15.9)

         ΔU = -1.1 10⁴ J = -0.11 10⁵ J

You can see that it is the same value.

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An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,
Taya2010 [7]

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Explanation:

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Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

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7 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

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