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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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We experience fictitious forces due to: a. Rotation of a reference frame b. Inertial reference frames c. Translational motion d.
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Answer:

A.

Explanation:

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The forces of inertia are, therefore, corrective terms to the real forces, which ensure that the formalism of Newton's laws can be applied unchanged to phenomena described with respect to a non-inertial reference system. The correct answer is A.

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
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Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

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Originally there must been

1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start

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