Work done against gravity to climb upwards is always stored in the form of gravitational potential energy
so we can say
here h = vertical height raised
so here we know that
here we have
now from above equation
so work done will be given by above value
In the question it is already given that the football player is 80 kg.
Then the mass of the football player = 80 kg
Velocity at which the football player is running = 8 m/s
<span>Kinetic Energy = 0.5 • mass • square of velocity
Now we have to put the known data in this equation to find the actual velocity of the footballer.
</span> <span></span>So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
= 0.5 * 80 * 64
= 2560
So the Kinetic energy of the footballer is 2560 joules
B represents the direction of the magnetic field around the wire
Explanation:
A wire carrying an electric current always produces a magnetic field around itself. The lines of the magnetic field produced by a current-carrying wires are concentric circles around the wire. The magnitude of the field is given by the formula:
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
The direction of the field lines is given by the so-called right hand rule, shown in the figure. Basically, the thumb of the right hand is placed in the direction of the electric current, while the other fingers are "wrapped" around the thumb: the direction of the other fingers give the direction of the magnetic field lines.
Learn more about magnetic field:
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Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ