Its been some time so i might be wrong but i think the answer is 3 either or 2
I made the drawing in the attached file.
I included two figures.
The upper figure shows the effect of:
- multiplying vector A times 1.5.
It is drawn in red with dotted line.
- multiplying vector B times - 3 .
It is drawn in purple with dotted line.
In the lower figure you have the resultant vector: C = 1.5A - 3B.
The method is that you translate the tail of the vector -3B unitl the point of the vector 1,5A, preserving the angles.
Then you draw the arrow that joins the tail of 1,5A with the point of -3B after translation.
The resultant arrow is the vector C and it is drawn in black dotted line.
Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone
f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N
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.
