<h3>Longitudinal Waves:</h3>
- Longitudinal waves are waves in which particles moves in the direction parallel to the direction of wave.
- Longitudinal wave are one directional waves.
- Sound waves and tension force applied on spring are examples of longitudinal waves.
<h3>Transverse Waves:</h3>
- Transverse waves in which particles move perpendicular to the direction of wave propagation.
- Transverse wave are two directional waves.
- Ripples formed on water and all electromagnetic waves are examples of transverse waves.
Explanation:
It is given that,
Earth's magnetic field, 
Direction of magnetic field, 67.1 degrees below the horizontal, with the horizontal component directed due north.
The magnetic force on the wire per unit length of wire, 
(a) The given scenario is shown in the attached figure as :
Using the right hand rule to find the direction of magnetic force on the wire. By using this rule, we get the magnetic force acting on the wire is in upward direction.
(b) Let I is the current flowing in the wire. The magnetic force is given by the following formula as:
I = 384.61 A
Hence, this is the required solution.
Answer:
Explained
Explanation:
1.Each of the spring scale will read 10N,considering acceleration due to gravity as 10 m/s^2
2.Each of the spring scale will read 10N because each string exerts a force of 10 N to counterbalance the force of 1 kg mass attached to it. This means the tension on the both side of the string is 10 N. So the scale will read 10 N. Also as spring balances are attached in series and kept on table so both spring balances will read same readings.
Answer: 
Explanation:
Given
Cross-sectional area 
Dielectric constant 
Dielectric strength 
Distance between capacitors 
Maximum charge that can be stored before dielectric breakdown is given by
![\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC](https://tex.z-dn.net/?f=%5CRightarrow%20Q%3DCV%5C%5C%5C%5C%5CRightarrow%20Q%3D%5Cdfrac%7Bk%5Cepsilon_oA%7D%7Bd%7D%5Ccdot%20%28Ed%29%5Cquad%5Cquad%20%5BV%3DE%5Ccdot%20d%5D%5C%5C%5C%5C%5CRightarrow%20Q%3Dk%5Cepsilon_oAE%5C%5C%5C%5C%5CRightarrow%20Q%3D4%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%200.4%5Ctimes%2010%5E%7B-4%7D%5Ctimes%202%5Ctimes%2010%5E8%5C%5C%5C%5C%5CRightarrow%20Q%3D28.32%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5C%5CRightarrow%20Q%3D283.2%5Ctimes%2010%5E%7B-9%7D%5C%20nC)
Answer:k = 10.83 N/m²
Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below
ω = √k/m
But ω = 2πf, where f = frequency.
f = number of oscillations /time taken
Number of oscillations = 14, time taken = 11s
f = 14/11 = 1.27Hz.
ω = 2×22/7×1.27
ω = 7.98 rad/s.
By substituting this parameters into ω = √k/m
Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.
7.98 = √k/0.17
By squaring both sides
(7.98)² = k/ 0.17
k = (7.98)² × 0.17
k = 10.83 N/m²
