Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s
Answer:B.As the north pole of the electromagnet nears the south pole of the permanent magnet, the current reverses and the poles of the magnets then repel.
Explanation:
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that there is no external torque on the system of two twins
so here we will use



Part b)
Since angular momentum is conserved here as there is no external torque
so we will have



Part c)
Work done by both of them = change in kinetic energy
so we have




Answer:
It is made up of molecules which are pulled down to Earth by gravity. That pull makes molecules bump into each other, exerting pressure. Our bodies are specially adapted to living under 1 kilogram per square centimeter (14.7 pounds per square inch) of pressure pushing down on us at sea level!
Explanation: