Answer:
A
Explanation:
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Answer:
The earth's gravitational force on the sun is equal to the sun's gravitational force on the earth
Explanation:
Newton's third law (law of action-reaction) states that:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
In other words, when two objects exert a force on each other, then the magnitude of the two forces is the same (while the directions are opposite).
In this problem, we can call the Sun as "object A" and the Earth as "object B". According to Newton's third law, therefore, we can say that the gravitational force that the Earth exerts on the Sun is equal (in magnitude, and opposite in direction) to the gravitational force that the Sun exerts on the Earth.
In physics, power is defined as energy per unit time. You will also hear it described as work per unit time. The standard unit of measure for power is the watt, where a watt is defined as joules (energy) per second (time). This is expressed as a fraction as J/s. If you wanted to increase the power in any operation, you can either increase the energy (more joules) or reduce the time (fewer seconds).
The Two examples of contact forces are:
- frictional force
- Contact force.
The two examples of non contact forces are:
- Gravitational force
- magnetic force.
Contact forces happens due to the contact between two objects
Non Contact forces happens because there is no contact between two objects. There is no attraction.
Answer:
Part a)
![F = 6 \times 10^{-3} N](https://tex.z-dn.net/?f=F%20%3D%206%20%5Ctimes%2010%5E%7B-3%7D%20N)
Direction of force is along the motion of charge
Part b)
![E = 4000 N/C](https://tex.z-dn.net/?f=E%20%3D%204000%20N%2FC)
direction of electric field is along the direction of motion
Explanation:
Part a)
As we know that the change in electric potential energy is equal to the work done by electric field
![W = EPE_A - EPE_B](https://tex.z-dn.net/?f=W%20%3D%20EPE_A%20-%20EPE_B)
![W = 9.0 \times 10^{-4} J](https://tex.z-dn.net/?f=W%20%3D%209.0%20%5Ctimes%2010%5E%7B-4%7D%20J)
now from the equation of work done we know that
![W = F.d](https://tex.z-dn.net/?f=W%20%3D%20F.d)
![(9.0 \times 10^{-4}) = F(0.15)](https://tex.z-dn.net/?f=%289.0%20%5Ctimes%2010%5E%7B-4%7D%29%20%3D%20F%280.15%29)
![F = 6 \times 10^{-3} N](https://tex.z-dn.net/?f=F%20%3D%206%20%5Ctimes%2010%5E%7B-3%7D%20N)
Direction of force is along the motion of charge
Part b)
As we know the relation between electrostatic force and electric field given as
![F = qE](https://tex.z-dn.net/?f=F%20%3D%20qE)
![(6 \times 10^{-3}) = 1.5 \times 10^{-6} E](https://tex.z-dn.net/?f=%286%20%5Ctimes%2010%5E%7B-3%7D%29%20%3D%201.5%20%5Ctimes%2010%5E%7B-6%7D%20E)
![E = 4000 N/C](https://tex.z-dn.net/?f=E%20%3D%204000%20N%2FC)
direction of electric field is along the direction of motion