Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
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
The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.
Answer:
Explanation:
We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.
The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.
Multiply.
1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.
<u>0.225 Newtons of force</u> are applied to the book.