Answer:
150156.25 Ω
Explanation:
Resistance: This can be defined as the opposition to the flow of electric current in a circuit. The S.I unit of resistance is Ohm's (Ω)
The expression for resistance is given as
P = V²/R................ equation 1
Where P = power, V = Voltage, R = Resistance.
Making R the subject of the equation,
R = V²/P.................. Equation 2
Given: V = 115 V , P = 0.16 W.
Substitute into equation 2
R = 155²/0.16
R = 150156.25 Ω
Hence,
The resistance = 150156.25 Ω
Lets say sphere 1 has a charge of 12 + and sphere 2 has a charge of 0 +. After they are touched Sphere 1 becomes 6 + and sphere 2 6 +. So 6 - 12 = a change of 6 -, while 6 - 0 = a change of 6 + Therfore,
Answer: The sign of the charge change / transfered are opposites.
Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved
Explanation:
Let the charges be q1 and q2 and the distance between the charges be 'd'
Mathematical representation of coulombs law will be;
F1=kq1q2/d²...(1)
Where k is the electrostatic constant.
If q1 and q2 is doubled and the distance halved, we will have;
F2 = k(2q1)(2q2)/(d/2)²
F2 = 4kq1q2/(d²/4)
F2 = 16kq1q2/d²...(2)
Dividing equation 1 by 2
F1/F2 = kq1q2/d² ÷ 16kq1q2/d²
F1/F2 = kq1q2/d² × d²/16kq1q2
F1/F2 = 1/16
F1 = 1/16F2
This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved
Answer: Radiation
Explanation: Radiation is the energy that comes from a source in form of electromagnetic waves, subatomic particles, light, or heat which travels through space.
Examples of radiation include the light, heat, and particles emitted from the Sun.
Using a foil barrier to prevent heat transfer is possible because foil has a silver color, and silver reflects light and heat instead of absorbing them. This is the opposite of black surfaces that absorb heat.
So in homes where these foil reflective barriers are used, the transfer of heat through Radiation is highly reduced.
Answer:
<em>Part A</em><em>:</em>
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
<em>Part B</em><em>:</em>
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
<em>Part C</em><em>:</em>
a) If the distance to the screen is decreased the fringe spacing will decrease.
<em>Part D</em><em>:</em>
The dot in the center of fringe E is
farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with


- m is the order number.
is the wavelength of the monochromatic light.- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light
is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be
We simply replace the values in that equation :


The dot in the center of fringe E is
farther from the left slit than from the right slit.