Answer:
E=930.84 N/C
Explanation:
Given that
I = 1150 W/m²
μ = 4Π x 10⁻⁷
C = 2.999 x 10⁸ m/s
E= C B
C=speed of light
B=Magnetic filed ,E=Electric filed
Power P = I A
A=Area=4πr² ,I=Intensity
E=930.84 N/C
Therefore answer is 930.84 N/C
The frequency will not change
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.
Answer:Same
Explanation:
Gravitational Potential energy is the function of height i.e. it is proportional to the height gained by an object or in other words work done against gravity is also gravitational Potential energy
For the same mass, if both attained the same height then gravitational Potential energy of both the friends is equal as it does not depend upon the path taken by the person.
<span>The mass of a 20 kg object is the same anywhere in the universe. That's its
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