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
I’m guessing it’s the last one, trough
Answer;
1 second
Explanation;
Two objects moving at the same speed will always stay the same distance apart. If two objects are moving at different speeds, the distance between them must change.
Therefore; if the distance will be the same and the speed is also the same then the time taken will be the same.
https://www.dailymotion.com/video/x4ug3zm
watch this video and u will get answer
Answer:
B = (μ₀*i/(2*π*x))*(x²-(a/2)²)/((b/2)²-(a/2)²)
Explanation:
Given
Outer diameter of the wire = b ⇒ R = b/2
Diameter of the clindrical hole at the center = a ⇒ r = a/2
The current that flows from left to right and is uniformly spread over the region between a and b = i
We apply Ampere's Law
Using the following formula for a/2 ≤ x ≤ b/2
B = (μ₀*i/(2*π*x))*(x²-(a/2)²)/((b/2)²-(a/2)²)
