An Olympic high diver has gravitational potential energy because of her height. As she dives, kinetic energy becomes of her energy just before she hits the water.
Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy that is related to gravitational force or to gravity.
Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles.
When the high diver is standing stable and not moving , that diver has a gravitational potential energy because of the height . The moment she dives , before hitting the water , from being stationary she gained some momentum and come in motion , due to motion her gravitational potential energy will change to kinetic energy before hitting the ground.
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Answer:
Yes
Explanation: Electric and magnetic field are known to be inter-related, this implies that for any current carrying conductor there is a resulting magnetic field around the wire ( for example a current carrying conductor deflects a compass) and a magnetic field has been known to produce some amount current based on the<em> </em>principle of electromagnetic induction by Micheal Faraday.
The strength of magnetic field generated by a current carrying conductor is given by Bio-Savart law (purely mathematical) which is
B =
B= strength of magnetic field
I =current on conductor
r = distance on any point of the conductor relative to it center
If a current carrying could generate this magnitude of magnetic field, thus this magnetic field has the ability to interact (exert a force on any magnetic material) with any other magnetic material including a magnet.
Yes, a current carrying conductor can exert a force on a magnetic field
Answer:
you have probably missed some details in the question.
Answer:
Initial velocity, U = 28.73m/s
Explanation:
Given the following data;
Final velocity, V = 35m/s
Acceleration, a = 5m/s²
Distance, S = 40m
To find the initial velocity (U), we would use the third equation of motion.
V² = U² + 2aS
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
35² = U + 2*5*40
1225 = U² + 400
U² = 1225 - 400
U² = 825
Taking the square root of both sides, we have;
Initial velocity, U = 28.73m/s
