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.
To learn more about Gravitational potential energy here
brainly.com/question/15978356
#SPJ4
The correct answer is C. Lost time is never found again
Explanation:
One of the most relevant personality theories is Type A and Type B personality theory that proposes two main types of personalities by grouping different personality traits. In the case of Type A personality, this refers to individuals that are very organized, impatient, and concerned with time and goals. On the opposite, Type B personality includes individuals that are more relaxed in different aspects.
According to this, one maxim that applies to Type A personality is "Lost time is never found again" because people with this personality are concerned about time and therefore, loss of time is considered highly negative by them. Also, due to their ambitions and concern with goals they want to avoid losing time as this is equivalent to work, money, goals, etc.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
The one that help create radio waves is :
Changing electric and magnetic fields applied at right angles
Radio waves are transverse wave, which means that the oscillations occurring perpendicular to the direction of energy transfer
hope this helps
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
