The formula for the period of wave is: wave period is equals to 1 over the frequency.
![waveperiod=\frac{1}{frequency}](https://tex.z-dn.net/?f=%20waveperiod%3D%5Cfrac%7B1%7D%7Bfrequency%7D%20)
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value
0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of
0.005 s.
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
<span> U = (1/2)kx^2
</span><span> U = (1/2)(5.3)(3.62-2.60)^2
</span> U = <span>
<span>2.75706 </span></span>J
Answer:
the distance between the submarine and the ocean floor is 11,250 m
Explanation:
Given;
speed of the wave, v = 1500 m/s
time of motion of the wave, t = 15 s
The time taken to receive the echo is calculated as;
![time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m](https://tex.z-dn.net/?f=time%20%5C%20of%20%5C%20motion%20%5C%20%28t%29%20%3D%20%5Cfrac%7Btotal%20%5C%20distance%20%7D%7Bspeed%20%5C%20of%20%5C%20wave%7D%20%3D%20%5Cfrac%7B2d%7D%7Bv%7D%20%20%5C%5C%5C%5C2d%20%3D%20vt%5C%5C%5C%5Cd%20%3D%20%5Cfrac%7Bvt%7D%7B2%7D%20%5C%5C%5C%5Cd%20%3D%20%5Cfrac%7B1500%20%5Ctimes%2015%7D%7B2%7D%20%5C%5C%5C%5Cd%20%3D%2011%2C250%20%5C%20m)
Therefore, the distance between the submarine and the ocean floor is 11,250 m
Answer:
Subduction, Trench, Mantle
Explanation:
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|= ![\sqrt{(8)^2+(-2)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%288%29%5E2%2B%28-2%29%5E2%7D)
|v|=
= 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.