To explain how transverse and longitudinal waves work, let us give two examples for each particular case.
In the case of transverse waves, the displacement of the medium is PERPENDICULAR to the direction of the wave. One way to visualize this effect is when you have a rope and between two people the rope is shaken horizontally. The shift is done from top to bottom. This phenomenon is common to see it in solids but rarely in liquids and gases. A common application usually occurs in electromagnetic radiation.
On the other hand in the longitudinal waves the displacement of the medium is PARALLEL to the direction of propagation of the wave. A clear example of this phenomenon is when a Slinky is pushed along a table where each of the rings will also move. From practice, sound waves enclose the definition of longitudinal wave displacement.
Therefore the correct answer is:
C. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.
Answer:
Explanation:
Given
Airplane is flying with horizontally with a constant momentum during time interval 
Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant
(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction
Answer:
Free body diagram
Explanation:
A free body diagram shows all the forces acting on a body. Since force is a vector quantity, the magnitude and direction of the forces are shown in the free-body diagram.
Force is the push or pull on a body.
It can be via contact or without contact. Such non-contact forces acts via a force field.
In physics, this free body diagram is used extensively.
Answer:
Explanation:
Left block is on surface with higher inclination so it will go down . If T be tension
For motion of block A ,
net force = mgsin60 - (T + mg cos 60 x μ ) , μ is coefficient of friction .
ma = mgsin60 - T - mg cos 60 x .1
10a = 277.13 - T - 16
= 261.13 - T
T = 261.13 - 10a
For motion of block B
T - mg sin30 - mgcos30 x μ = ma
T- 160 - 27.71 = 10 a
261.13 - 10a - 160 - 27.71 = 10a
73.42 = 20a
a = 3.67 ft / s²
common acceleration = 3.67 ft / s²
Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa
