Determine whether or not each of the following statements is true. If a statement is true, prove it. If a statement is false, pr
1 answer:
You might be interested in
Answer:
<u>r < a:</u>
<u>r = a:</u>
<u>a < r < b:</u>
<u>r = b:</u>
<u>b < r < c:</u>
E = 0
<u>r = c:</u>
<u>r < c:</u>
Explanation:
Gauss' Law will be applied to each region to find the E-field.
An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.
<u>r<a:</u>
Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;
Applying Gauss' Law:
The minus sign determines the direction of the field, which is towards the center.
<u>At r = a: </u>
<u>At a < r < b:</u>
The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.
<u /><u />
<u>At r = b:</u>
<u /><u />
Again, the minus sign indicates the direction of the field towards the center.
<u>At b < r < c:</u>
The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.
Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.
E = 0.
<u>At r < c:</u>
The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).
<u>At r = c:</u>
A. IMA: 4
The Ideal Mechanical Advantage (IMA) is given by:
where
is the input distance
is the output distance
For the pulley system in this problem, and
, so the IMA is
B. MA: 3.59
The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by
where is the output force and
is the input force. For the pulley system in this problem,
and
, so the MA is
C. Efficiency: 89.8 %
The efficiency of a machine is equal to the ratio between the MA and the AMA:
Therefore, in this case,
Answer:
225 rpm
Explanation:
The angular acceleration of the fan is given by:
where
is the final angular speed
is the initial angular speed
is the time interval
For the fan in this problem,
Substituting,
Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:
where
Substituting,