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.
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.
First of all, we need to calculate the speed of sound at temperature of

:

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):

where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
Answer: 35*10^3 N/m
Explanation: In order to explain this problem we know that the potential energy for spring is given by:
Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.
We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.
Then the difference of energy for both cases is 7 J so:
ΔUp= 1/2*k* (0.02)^2 then
k=2*7/(0.02)^2=35000 N/m
If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.