Answer:
P.E = 0.068 J = 68 mJ
Explanation:
First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = -9.8 m/s² (negative sign due to upward motion)
h = height attained by the ball toy = ?
Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)
Vi = Initial Velocity = 3 m/s
Therefore,
2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²
h = (9 m²/s²)/(19.6 m/s²)
h = 0.46 m
Now, the gravitational potential energy of ball at its peak is given by the following formula:
P.E = mgh
P.E = (0.015 kg)(9.8 m/s²)(0.46 m)
<u>P.E = 0.068 J = 68 mJ</u>
Answer:
Gravity.” or “Gravity causes the moon to orbit the earth.” “Acceleration due to gravity” is much more precise. It is the acceleration that an object experiences when it is dropped with no other external forces on it. ... Near the surface of the earth the acceleration due to gravity is about 9.8
The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
<h3>What is an electric field?</h3>
An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
The given data in the problem is given by;
E is the electric field = (200 N/C)
d is the distance = 5.0 cm.=0.05 m
Q is the charge of electrons= 1.602 x 10^-19 C
The formula for electric potential is given by;
The work is defined as the product of the potential difference and charge of an electron.
Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
To learn more about the electric field refer to the link;
brainly.com/question/15071884
Answer:
By induction method
Explanation:
Induction method involves charging an electrically neutral body by bringing it in contact with an electrically charged body.
For the electrophus, a charge opposite that on the slab is induced on the side in contact with the slab; driving the opposite charge (this will be the same as that on the slab) to the other end of the elctrophus. Touching the electrophus removes the charge opposite the charge induced on the electrophus by the charged slab either by drawing up charge from the earth or taking the charge to earth (depends on the charge. A negative charge is drawn to earth while a positive charge draws up electrons from the earth)
