Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
Answer:
6538.8 Angstrom
Explanation:
work function, w = 1.9 eV = 1.9 x 1.6 x 10^-19 J = 3.04 x 10^-19 J
Let the longest wavelength is λ.
W = h c / λ
λ = h c / W
λ = (6.626 x 10^-34 x 3 x 10^8) / (3.04 x 10^-19)
λ = 6.5388 x 10^-7 m = 6538.8 Angstrom
Thus, the longest wavelength is 6538.8 Angstrom.
Answer:
Constructive interference
Explanation:
- This is an example of a standing wave produced when two ends of a string are oscillated in the same plane. The displacement of of point on two ends oscillates vertically.
- We are given that two pulses move along the string each coming towards each other and meet at a common point ( P ).
- Each pulse have their own magnitude or displacement in the vertical plane. If the pulses are to meet at a common point at the same instant, then they interfere with each other constructively.
- Where constructive interference of two pulses is the addition of magnitudes of induvidual pulses and form a single puls of the constructed magnitude.
magnitude ( New pulse ) = magnitude (Pulse 1) + magnitude (Pulse 2)
Answer: I think your answer would be true.
Answer:
0.36
Explanation:
The maximum force of friction exerted by the surface is given by:
(1)
where
is the coefficient of friction
N is the normal reaction
The shed's weight is 2200 N. Since there is no motion along the vertical direction, the normal reaction is equal and opposite to the weight, so
N = 2200 N
The horizontal force that is pushing the shed is
F = 800 N
In order for it to keep moving, the force of friction (which acts horizontally in the opposite direction) must be not greater than this value. So the maximum force of friction must be
And substituting the values into eq.(1), we can find the maximum value of the coefficient of friction:
