Answer:
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is
Speed of nucleus at A is
Potential at point A,
Potential at point B,
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
So, the speed at point B is .
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:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.