Answer:
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
Explanation:
Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution
As we know that the distance moved in one revolution is given as
also the time period of revolution for both will remain same as they move with the time period of carousel
Now we can say that the speed is given as
so Juan will have less tangential speed. so correct answer will be
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
Answer:
m = mass of the penny
r = distance of the penny from the center of the turntable or axis of rotation
w = angular speed of rotation of turntable
F = centripetal force experienced by the penny
centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as
F = m r w²
in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .
hence greater the distance from center , greater will be the centripetal force to remain in place.
So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.
Explanation:
3 trams must be added
Explanation:
In this problem, there are 12 trams along the ring road, spaced at regular intervals.
Calling L the length of the ring road, this means that the space between two consecutive trams is
(1)
In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become
And the number of trams will become
So eq.(1) will become
(2)
And substituting eq.(1) into eq.(2), we find:
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Answer:
(1) En to n-1 = 0.55 ev
(2) En-1 to n-2 = 0.389 ev
(3) ninitial =4
(4) L =483.676 ×10^-11 nm
(5) λlongest= 1773.33 nm
Explanation:
Detailed explanation of answer is given in the attached files.
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.
