Answer: D
Explanation: Look at the figure, we can conclude that the correct answer is
X: Low potential energy
Y: High potential energy
Z: Flow of electrons
Because electrons flow where there is difference in potential energy. And electrons move from a region of high potential energy to a region of low potential energy.
Since the arrow is pointing to X, that means
Z is the flow of electrons, X is of low potential energy and Y is of high potential energy.
Answer:
2.5 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s²
For circular motion, the expression for acceleration is given as,
a = ω²r ................ Equation 1
Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.
Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.
Substitute into equation 1
a = 5²(0.10)
a = 25(0.10)
a = 2.5 m/s²
Hence the acceleration of the particle = 2.5 m/s²
Hence, none of the option is correct
The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
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Answer:

Explanation:
P = Power Output = 1000 W
r = Radius = 35000000 m
= Permittivity of free space = 
c = Speed of light = 
Intensity of Electric radiation is given by

Intensity of Electric radiation is given by

The amplitude of the electric field vector is 
To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables
Yield Strength of the metal specimen

Yield Strain of the Specimen

Diameter of the test-specimen

Gage length of the Specimen

Modulus of elasticity



Strain energy per unit volume at the elastic limit is



Considering that the net strain energy of the sample is




Therefore the net strain energy of the sample is 