The electrical force acting on a charge q immersed in an electric field is equal to
where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
Weight on moon = (0.16) • Earth weight
Answer:
4.6s
Explanation:
v=u+at
0=22.5+(-9.8)t
-22.5=-9.8t
t=-22.5/-9.8
t=2.295 s
The total time will double
2.295×2=4.59s
=4.6s
Answer:
Yes, there a link between number of bulbs and current drawn from the power pack.
Explanation:
In an Electrical circuit, we have resistors present in that circuit. These resistors can be connected in two ways.
a) Series connection
b) Parallel connection
There is a link or a relationship between number of bulbs and the current drawn from the power pack. This is because the number of bulbs is equivalent to or equal to the number of resistors.
Hence,
a) In a series connection, the link or relationship between the number of bulbs(resistors) is as the number of light bulbs increases, the current in the power pack (circuit) decreases.
b) In a parallel connection, the link or relationship between the number of bulbs(resistors) is as the number of light bulbs increases, the current in the power pack (circuit) increases.
Answer:
a) laser 1 has the maximum closest to the central maximum
b) y₂ –y₁ = L 1.66 10⁻²
Explanation:
a), B1, B2) The expression that describes the constructive interference for a double slit is
d sin θ = m λ
The pattern is observed on a screen
tan θ = y / L
Since the angles are very small
tan θ = sin θ / cos θ = sin θ = y/L
d y / L = m λ
In this case the laser has a wavelength
λ
₁ = d/20
We substitute
d y / L = m d / 20
m = 1
y₁ = L / 20
For the laser 2 λ
₂= d / 15
y₂ = L / 15
When examining the two expressions, laser 1 has the maximum closest to the central maximum
b) the difference between the two patterns is
y₂- y₁ = L (1/15 - 1/20)
y₂ –y₁ = L 1.66 10⁻²
C) laser 1 second maximum
y₁ ’= 2 L / 20
y₁ ’= L 0.1
Laser 2 third minimum
To have a minimum, the equation must be satisfied
d sin θ = (m + ½) λ
d y / L = (m + ½) λ
d y / L = (m + ½) d / 15
y = L (m +1/2) / 15
m = 3
y₂’= L (3 + ½) / 15
y₂’= L 0.2333
The difference is
y₁ ’- y₂’ = L (0.1 - 0.2333)
y₁ ’–y₂’ = L (-0.133)
