Answer:
<em>The glider's new speed is 68.90 m/s</em>
Explanation:
<u>Principle Of Conservation Of Mechanical Energy</u>
The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

The gravitational potential energy when it's at a height h from the zero reference is

The total mechanical energy is


The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.
The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

Operating in terms of m


Then we know the glider dives to 278 meters and we need to know their final speed, let's call it
. The final mechanical energy is

Operating and factoring

Both mechanical energies must be the same, so

Simplifying by m and rearranging

Computing

The glider's new speed is 68.90 m/s
Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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