Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data
Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)
The increase in potential energy of the ball is 115.82 J
To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that
Where,
N = Number of loops
A = Cross sectional Area
B = Magnetic Field
Therefore the correct answer is A.
If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.
A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC, is on the x axis at x=0.500 m.
No, there is no point at a finite distance where the electric potential is zero.
Hence, Option D) is correct.
What is electric potential?
Electric potential is the capacity for doing work. In the electrical case, a charge will exert a force on some other charge and the potential energy arises. For example, if a positive charge Q is fixed at some point in space, any other positive charge when brought close to it will experience a repulsive force and will therefore have potential energy.
It is also defined as the amount of work required to move a unit charge from a reference point to a specific point against an electric field.
To learn more about electric potential, refer to:
brainly.com/question/15764612
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a. We can calculate the amount of work by calculating the area under the graph.
first area (rectangular): 2.5 x 6 = 15
second area(trapezoid): 1/2 x (6+10) x 2.5 =20
total work done: 35 J
b. the force was first applied = 6 N
F = m.a
a = 6 : 3 = 2 m/s²
vf²=vi²+2as
vf²=6²+2.2.5
vf²=56
vf=7.5 m/s
