Transverse Waves: Displacement of the medium is perpendicular to the direction of propagation of the wave.
To understand this it is good to think of a rope being held still by person B and being moved up and down by person A. The direction of propagation is from person A to B, so you will see the waves move along this way. But the displacement will be up and down.
Can travel in solids, but not in liquids and gas.
eg. Electromagnetic radiation
Longitudinal Waves: Displacement of the medium is parallel to the direction of propagation of the wave.
A good example for this is a slinky being pushed along the table, the propagation will be along the table and so will the displacement of all the 'rings'.
Can travel through all states of matter.
eg. Sound waves
We know from Newtons law of force that
Work done = Force * Distance moved
We know from the question that
Force applied = 70 Newton
Distance moved = 9 meters.
Since the object is moving in the same direction as the force applied
So
Work done = Force * Distance moved
= 70 * 9 newton meters(Nm)
= 630 newton meters(Nm)
= 630 joules
Answer:
4.96 km/hr
Explanation:
Given that,
A person is running west in Moscow at 2.1 km/hr relative to the ground
Another person is jogging east at 4.5 km/hr.
We need to find the velocity of the jogger relative to the ground. Using the concept of relative speed, let V is the required velocity. So,
So, the velocity of the jogger relative to the ground is 4.96 km/hr.
Answer:
Explanation:
λ=c x²
c = λ / x²
λ is mass / length
so its dimensional formula is ML⁻¹
x is length so its dimensional formula is L
c = λ / x²
= ML⁻¹ / L²
= ML⁻³
B )
We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M
The mass in the rod is symmetrically distributed on both side of middle point.
we consider a small strip of rod of length dx at x distance away from middle point
its mass dm = λdx = cx² dx
By integrating it from -L to +L we can calculate mass of whole rod , that is
M = ∫cx² dx
= [c x³ / 3] from -L/2 to +L/2
= c/3 [ L³/8 + L³/8]
M = c L³/12
c = 12 M L⁻³
C ) Moment of inertia of rod
∫dmx²
= ∫λdxx²
= ∫cx²dxx²
= ∫cx⁴dx
= c x⁵ / 5 from - L/2 to L/2
= c / 5 ( L⁵/ 32 +L⁵/ 32)
= (2c / 160)L⁵
= (c / 80) L⁵
= (12 M L⁻³/80)L⁵
= 3/20 ML²
=
=