The second diver have to leap to make a competitive splash by 4.08 m high.
<h3>What is potential energy?</h3>
The energy by virtue of its position is called the potential energy.
PE = mgh
where, g = 9.81 m/s²
Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.
The potential energy of the first diver must be equal to the second diver.
P.E₁ = P.E₂
m₁gh₁ = m₂gh₂
Substitute the vales, we have
136 x 3 = 100 x h₂
h₂ = ₂4.08 m
Thus, the second diver need to leap by 4.08 m high.
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Answer:
The work done by a particle from x = 0 to x = 2 m is 20 J.
Explanation:
A force on a particle depends on position constrained to move along the x-axis, is given by,
We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.
We know that the work done by a particle is given by the formula as follows :
So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Force = mass × acceleration
To find acceleration, we can divide the speed by the time it took:
acceleration = 2.40×10^7 / 1.8×10^-9
acceleration = 1.33×10^16
the mass is equal to the mass of an electron
force = (9.11×10^-31)(1.33×10^16)
force = 1.21×10^-14 N
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
<h3>What do you mean by electric potential? </h3>
The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.
V = kq/r
Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.
Electric potential at O due to four charges is given by,
V = 4KQ/ r
where, r = √2a/2 = a/√2
V = 4k × Q√2/a
V = √2Q/πÈa
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
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