Answer:
An <u>applied force</u> is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.
A <u>friction force</u> is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:
= µ • 
Answer: c living in a camber in an under water habitat
Explanation:
Answer:
2. Move faster
Explanation:
Because you lighten the weight and pushed at the same speed it is easier to push the 400-grams than the 800-grams.
Have a wonderful day!
Answer:
So coefficient of kinetic friction will be equal to 0.4081
Explanation:
We have given mass of the block m = 0.5 kg
The spring is compressed by length x = 0.2 m
Spring constant of the sprig k = 100 N/m
Blocks moves a horizontal distance of s = 1 m
Work done in stretching the spring is equal to 
This energy will be equal to kinetic energy of the block
And this kinetic energy must be equal to work done by the frictional force
So 
So coefficient of kinetic friction will be equal to 0.4081
Answer:
12 nC
Explanation:
Capacity of the parallel plate capacitor
C = ε₀ A/d
ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate
Area of plate = π r²
= 3.14 x (0.8x 10⁻²)²
= 2 x 10⁻⁴
C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³
= 7.08 x 10⁻¹³
Potential difference between plate = field strength x distance between plate
= 6 x 10⁶ x 2.8 x 10⁻³
= 16.8 x 10³ V
Charge on plate = CV
=7.08 x 10⁻¹³ X 16.8 X 10³
11.9 X 10⁻⁹ C
12 nC .
