On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.
If the rock starts out moving upward at 10 m/s, then it will
continue upward for only (10/9.8) = 1.02 second, before
it stops rising and starts falling.
Its average speed during that time is (1/2) (10 + 0) = 5 m/s .
At an average speed of 5 m/s for 1.02 sec,
the rock rises
(5 m/s) x (1.02 sec) = 5.102 meters .
Given the following in the problem:
Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s
Get the wavelength of the sound by using the formula:
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2
Get the path length difference to the point from the two speakers
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47
x = delta / lambda
If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.
Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference.
Step 1: list known info
distance(change in position (Δx))= 18m+22m= 40m
time= 20 seconds
Step 2 :solve for velocity
velocity= Δx÷time
v= 40/20= 2m/s
Answer: the velocity is 2 meters per a second (m/s)
Answer:
A) 
B) 
C) 
D) mosquitoes speed in part B is very much larger than that of part C.
Explanation:
Given:
- Distance form the sound source,

- sound intensity level at the given location,

- diameter of the eardrum membrane in humans,

- We have the minimum detectable intensity to the human ears,

(A)
<u>Now the intensity of the sound at the given location is related mathematically as:</u>
..........................................(1)



<em>As we know :</em>


is the energy transferred to the eardrums per second.
(B)
mass of mosquito, 
<u>Now the velocity of mosquito for the same kinetic energy:</u>



(C)
Given:
- Sound intensity,

<u>Using eq. (1)</u>



Now, power:



Hence:




(D)
mosquitoes speed in part B is very much larger than that of part C.
Assuming that reaching a height 0 doesn’t stop the ball, and that it accelerates at 9.8 m/s^2, the ball would be traveling at 0.5 + 0.7*9.8 = 7.36 m/s downwards.