<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>
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N
Answer:
<h2>62.5 m/s</h2>
Explanation:
The speed of the car can be found by using the formula
d is the distance
t is the time
From the question we have
We have the final answer as
<h3>62.5 m/s</h3>
Hope this helps you
Answer:
.737 v
Explanation:
Since they are in series....they all have the same current running through them.....find the total resistance to calculate the current:
R = 67 + 83 + 433 + 309 = 892 ohm
V/R = current = 7.92 / 892 = 8.87 mAmps
Now the voltage across ecah resistor is I R
for the second one 8.87 ma * 83 ohm = V = .737 V
