At a playground a student runs at a speed of 5 m/s and jump onto a circular disk of radius 3/2 m that is free to rotate around a
1 answer:
Answer:
e. 4 rad/s
Explanation:
The maximum amount of angular velocity that can be reached by the disk can be found using the following formula:
where,
ω = maximum angular velocity = ?
v = linear speed = 5 m/s
r = radius of disc = 1.5 m
Therefore,
Therefore, the angular speed can be lower than this value due to frictional losses, but it can not exceed this value in the given linear speed limit of 5 m/s.
Hence, the correct option is:
<u>e. 4 rad/s</u>
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Answer:
see explanation
Explanation:
You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:
See the attached table.
From the left we have:
r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min
From the right we have:
r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.
And this should be the correct answer. Watch your chart and replace if it's neccesary.
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).
Learn more about Gaussian sphere here:
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