Answer:
Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.
Explanation:
So, we are given the following set of infomation in the question given above;
=> "spherical Gaussian surface of radius R centered at the origin."
=> " A charge Q is placed inside the sphere."
So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?
The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;
REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.
Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)
Well first of all, the Space Shuttle program ended a few years ago, and none have been launched since then.
The Shuttle never went to places that were properly referred to as "outer space". When they flew, the Space Shuttles went to low Earth orbit, where the acceleration of gravity is roughly 85% of its value on the Earth's surface.
So a Shuttle that weighed 20 million Newtons on the launch pad weighed roughly 17 million Newtons while in orbit.
Answer:
It will take Andy 1.198minutes to mow the lawn and it will take Brian 1,209.98minutes to mow the same lawn.
Step-by-step explanation:
Using the simultaneous equation concept, let A denote Andy and B be Brian
Andy and Brian can mow the lawn for 1212minutes i.e A+B = 1212..eqn 1
If Brian would mow the lawn by himself in 1010 minutes more than it would take Andy, this means B=1010A...eqn 2.
Substituting eqn 2 into eqn 1
Equation 1 becomes
A+1010A=1212
1011A=1212
A=1212/1011
A=1.198
B = 1010×1.198
B=1,209.98
Therefore, It will take Andy 1.198minutes to mow the lawn and it will take Brian 1,209.98minutes to mow the same lawn.
