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Komok [63]
3 years ago
14

PLEASE HELPPPP ME WITH THIS

Physics
1 answer:
lozanna [386]3 years ago
4 0
I am not so sure about this it is too difficult
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Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. the tide measures 5.15 ft at
RSB [31]
<span>f(x) = 5.05*sin(x*pi/12) + 5.15

   First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

   So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
  f(x) = 5.05*sin(x*pi/12) + 5.15

   The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
 f(x) = 5.05*sin(x*pi/12 + C) + 5.15
  where
 C = Phase correction offset.

   But we don't need it for this problem, so the answer is:
 f(x) = 5.05*sin(x*pi/12) + 5.15

   Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
7 0
3 years ago
If
timama [110]

Answer:

Workdone = 1960 Joules.

Explanation:

Given the following data;

Mass = 5kg

Force = 49N

Height (distance) = 40m

To find the workdone;

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 49*40

Workdone = 1960 Joules.

Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.

7 0
2 years ago
A car dropped from a height of 44 meters fall to a height of zero meters. How fast will the car be traveling as it hits the grou
FinnZ [79.3K]
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed

Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
Vf=29.3m/s
6 0
2 years ago
Quick Physics Question
sveticcg [70]
It is Real,Virtual,The Same Size, Inverted
8 0
2 years ago
A test pilot flies with an acceleration of of 5 g . what is the acceleration in meter per second
JulijaS [17]
I mean if he flies 5g that means that's his average speed too
5 0
3 years ago
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