Answer
Force between two charged bodies q₁ and q₂ is given by the formula below.
F = Kq₁q₂/r²
Where k = constant
r = distance between them
F = (9.0 E9 × 1.0 E-6 × 1.0 E-6)/0.30²
= 9.0E-3/0.30²
= 9.0E-3/0.09
= 0.1 N
Answer:
3 m
Explanation:
At last, Same line; so, 6-3 = 3
P = m/v
6000 kg/m3 = m/4x10^-4
m = 2.4 kg
P = m/v
5000 kg/m3 = 2.4kg/v
V = 4.8 x 10^-4
0.7N x 10 = 7kg
P = m/v
P = 7/4.8 x 10^-4
P answer = 1458.3
It is right?
Answer:
0.5639m
Explanation:
For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ
=sin⁻¹(mλ/d)
mλ /d =y/L
for the first order,
y= mλL/d
For ratio separation y₀/yD=1 and d= 1
y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]
1=λ ₀L₀/λD.LD.
λD.LD= λ ₀L₀
L₀= λD.LD/ λ ₀..............(1)
Then substitute the given values into (1) we have
L₀=471 *0.497/611
= 0.3831m
Distance by which the screen has to be moved towards the slit is
LD- Lo
0.947-0.3831= 0.5639m
Answer:
Explanation:
Given that
Mass of disc,M=3 kg
radius r= 65 cm
Mass of small mass ,m=0.07 kg
Initial speed = 2.2 rev/s
If external torque is zero then angular momentum of system will remain conserve.
Moment of inertia of disc at initial condition
Moment of inertia of disc at final condition
So from conservation of angular momentum
This is final speed of disc after small mass flies off from the disc.