The frequency of the wave has not changed.
In fact, the frequency of a wave is given by:
where v is the wave's speed and 
is the wavelength.
Applying the formula:
- In air, the frequency of the wave is:
- underwater, the frequency of the wave is:
So, the frequency has not changed.
A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.
Answer: Option D
<u>Explanation:</u>
Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.
There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ
= σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23
Answer:
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Explanation:
Magnetic flux is the scalar product of the magnetic field over the area
Ф = ∫ B. dA
where B is the magnetic field and A is the area
Let's look at stationary, for which factors affect flow
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
