let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer: 
Explanation:
Given
mass of laptop m=2 kg
The velocity of car u=8 m/s
The coefficient of static friction is 
The coefficient of kinetic friction is 
As the car is moving, so the coefficient of kinetic friction comes into play
deceleration offered by friction 
Using the equation of motion 
insert the values

Answer:
Explanation:
energy emitted by source per second = .5 J
Eg = 1.43 eV .
Energy converted into radiation = .5 x .12 = .06 J
energy of one photon = 1.43 eV
= 1.43 x 1.6 x 10⁻¹⁹ J
= 2.288 x 10⁻¹⁹ J .
no of photons generated = .06 / 2.288 x 10⁻¹⁹
= 2.6223 x 10¹⁷
wavelength of photon λ = 1275 / 1.43 nm
= 891.6 nm .
momentum of photon = h / λ ; h is plank's constant
= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹
= .0074 x 10⁻²⁵ J.s
Total momentum of all the photons generated
= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷
= .0194 x 10⁻⁸ Js
b ) spectral width in terms of wavelength = 30 nm
frequency width = ?
n = c / λ , n is frequency , c is velocity of light and λ is wavelength
differentiating both sides
dn = c x dλ / λ²
given dλ = 30 nm
λ = 891.6 nm
dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²
= 11.3 x 10¹² Hz .
c )
10 nW = 10 x 10⁻⁹ W
= 10⁻⁸ W .
energy of 50 dB
50 dB = 5 B
I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s
I = I₀ x 10⁵
= 10⁻¹² x 10⁵
= 10⁻⁷ W
= 10 x 10⁻⁸ W
power required
= 10⁻⁸ + 10 x 10⁻⁸ W
= 11 x 10⁻⁸ W.
Answer:

where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have


since string makes 30 degree angle with the vertical so we will have





where E = electric field intensity
Answer:
E= 4.35*10^6 N/C
Explanation:
Let's find the area charge density of the plate
α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2
Now we can calculate the electric field just of the plate
E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C