At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the oppo
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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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Answer:
The resultant velocity of the helicopter is .
Explanation:
Physically speaking, the resulting velocity of the helicopter (), measured in meters per second, is equal to the absolute velocity of the wind (
), measured in meters per second, plus the velocity of the helicopter relative to wind (
), also call velocity at still air, measured in meters per second. That is:
(1)
In addition, vectors in rectangular form are defined by the following expression:
(2)
Where:
- Magnitude, measured in meters per second.
- Direction angle, measured in sexagesimal degrees.
Then, (1) is expanded by applying (2):
(3)
If we know that ,
,
and
, then the resulting velocity of the helicopter is:
The resultant velocity of the helicopter is .