So picture below to see answer and explanation.
Answer:
a). The potential is highest at the center of the sphere
Explanation:
We k ow the potential of a non conducting charged sphre of radius R at a point r < R is given by
![E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-(\frac{r}{R})^{2} \right ]](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7BK.Q%7D%7B2R%7D%20%5Cright%20%5D%5Cleft%20%5B%203-%28%5Cfrac%7Br%7D%7BR%7D%29%5E%7B2%7D%20%5Cright%20%5D)
Therefore at the center of the sphere where r = 0
![E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-0 \right ]](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7BK.Q%7D%7B2R%7D%20%5Cright%20%5D%5Cleft%20%5B%203-0%20%5Cright%20%5D)
![E=\left [ \frac{3K.Q}{2R} \right ]](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7B3K.Q%7D%7B2R%7D%20%5Cright%20%5D)
Now at the surface of the sphere where r = R
![E=\left [ \frac{K.Q}{2R} \right ]\left ( 3-1 \right )](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7BK.Q%7D%7B2R%7D%20%5Cright%20%5D%5Cleft%20%28%203-1%20%5Cright%20%29)
![E=\left [ \frac{2K.Q}{2R} \right ]](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7B2K.Q%7D%7B2R%7D%20%5Cright%20%5D)
![E=\left [ \frac{K.Q}{R} \right ]](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7BK.Q%7D%7BR%7D%20%5Cright%20%5D)
Now outside the sphere where r > R, the potential is
![E=\left [ \frac{K.Q}{r} \right ]](https://tex.z-dn.net/?f=E%3D%5Cleft%20%5B%20%5Cfrac%7BK.Q%7D%7Br%7D%20%5Cright%20%5D)
This gives the same result as the previous one.
As ![r\rightarrow \infty , E\rightarrow 0](https://tex.z-dn.net/?f=r%5Crightarrow%20%5Cinfty%20%2C%20E%5Crightarrow%200)
Thus, the potential of the sphere is highest at the center.
Forward and downward speed are referred to as acceleration and deceleration respectively.
<h3>
What is Acceleration?</h3>
This is the rate at which an object gains speed with respect to time with its unit being meter per second square(m/s²).
Objects in constant speed are said to have a uniform motion. The downward speed is known as deceleration which involves the object losing speed to return to a state of rest.
Read more about Acceleration here brainly.com/question/460763
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Answer:
c(t)= 40t
Explanation:
As it forms the right angle triangle. see attachment for the figure.
We will use Pythagoras theorem i.e
a² + b² = c²
where,
'a' is perpendicular
'b' is base
'c' is hypotenuse i.e the distance D between the ships in terms of the time 't'
the distance 'a' from the starting point is 24t
the distance 'b' from the starting point is 32t
therefore,
(24t)²+ (32t)² = c(t)²
c(t)= √(24t)²+ (32t)²
c(t)= √576t² + 1024t²
c(t)=√ 1600t²
c(t)= 40t
Thus, function that models the distance D between the ships in terms of the time t (in hours) elapsed since their departure is 40t