In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
The wavelength of the 2nd harmonic is:
The wavelength of the 4th harmonic is:
It is not possible to find any integer n such that 
, therefore the correct options are A, B and D.
Answer:
A few of the positive particles aimed at a gold foil seemed to bounce back
Explanation:
Answer: critical angle, sin^-1 (n2/n1)
Explanation: the angle of incidence at which the retracted ray makes an angle of 90° with the normal is known as the critical angle.
Snell's law defined refraction mathematically as shown below
n1 sin θi = n2 sin θr
n1 = refractive index of the first medium
n2 = refractive index of the second medium
θi = angle of incidence
θr = angle of refraction
When the refrafted ray is perpendicular to the normal, the angle of refraction (θr) is 90° hence making the angle of incidence (θi) the critical angle θc
By substituting these conditions into the Snell's law, we have that
n1 sin θc = n2 sin 90
According to trigonometry, the value of sin 90 is 1, hence we have that
n1 sin θc =n2
sin θc = n2/n1
θc = sin^-1 (n2/n1)
Solution :
Let 
kg
m/s
Let 
and 
are the speeds of the disk 
and 
after the collision.
So applying conservation of momentum in the y-direction,
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
m/s
So, 
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
