Answer:
1.424 μC
Explanation:
I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),
tension T (acting along the string - to the pivot point), and
F (electric force – acting along the line connecting the charges).
We then have something like this
x: T•sin α = F,
y: T•cosα = mg.
Dividing the first one by the second one we have
T•sin α/ T•cosα = F/mg, ultimately,
tan α = F/mg.
Since we already know that
q1=q2=q, and
r=2•L•sinα,
k=9•10^9 N•m²/C²
Remember,
F =k•q1•q2/r², if we substitute for r, we have
F = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • √(m•g•tanα/k)=
=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =
q = 0.486 • √(8.61•10^-12)
q = 0.486 • 2.93•10^-6
q = 1.424•10^-6 C
q = 1.424 μC.
I am sure the ANSWER is A !!
Answer:
Explanation:The rate of decrease of velocity is given with respect to displacement. As v approaches zero when s approaches 39, it can be written as
where c=19
when s=8 we have
hence a can be found by the chain rule
From the equation
the s in term of t can be found
Hence it can be seen that if s approaches 39 the t function becomes undefined hence s is never 39m
B.
increases as the tension of the string increases
Answer:
Explanation:
It is given that,
Mass of the loop, m = 10 g = 0.01 kg
Resistance of the loop, R = 0.02 ohms
Dimension of square loop, 7 cm × 7 cm
Area of the loop,
At time t = 0 s, the magnetic field increases from 0 to 1 T in 0.01 s
(a) Due to change in magnetic field, an emf is induced in the loop. Using the formula for induced emf as :
Now using Ohm's law to find the induced current in the loop. It is given by :
I =24.5 A
(b) A magnetic force acting on the loop is given by :
F = 0.8575
Since,
v = 0.8575 m/s
Hence, this is the require solution.