The restriction are that neither h nor g may equal zero as that would lead to division by zero.
The simplification after units cancel is:
(3(gh)^3)/100 or if you prefer
0.03(gh)^3
Answer:
The plane's distance from the radar station will increase about 8 miles per minute when it is 5 miles away from it.
Step-by-step explanation:
When the plane passes over the radar station, the current distance is the altitude h = 2. Then it moves b horizontally so that the distance to the station is 5. We can form a rectangle triangle using b, h and the hypotenuse 5. Therefore, b should satisfy
h²+b² = 5², since h = 2, h² = 4, as a result
b² = 25-4 = 21, thus
b = √21.
Since it moved √21 mi, then the time passed is √21/540 = 0.008466 hours, which is 0.51 minutes. Note that in 1 minute, the plane makes 540/60 = 9 miles.
The distance between the plane and the radar station after x minutes from the moment that the plane passes over it is given by the function
We have to compute the derivate of f in x = 0.51. The derivate of f is given by
also,
The plane's distance from the station will increase about 8 miles per minute.
Answer:
C - 13*
Step-by-step explanation:
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C