Answer:
I believe the answer is D.
Explanation:
Protons are found inside the nucleus so are neutrons. Electrons are found outside the nucleus.
Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :

(c) Moment of inertia :
The net torque acting on it is,
, I is the moment of inertia
Also, 
So,

Hence, this is the required solution.
Let say the point is inside the cylinder
then as per Gauss' law we have

here q = charge inside the gaussian surface.
Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.
we will calculate the charge first which is given as


now using the equation of Gauss law we will have


now we will have

Now if we have a situation that the point lies outside the cylinder
we will calculate the charge first which is given as it is now the total charge of the cylinder


now using the equation of Gauss law we will have


now we will have
It produces only virtual images is the answer
Answer:

Explanation:
To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.
By conservation of energy we know that,

Where,

Replacing


Our values are given by,





Replacing at the equation,


Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was 