Answer:
Static Friction - acts on objects when they are resting on a surface
Sliding Friction - friction that acts on objects when they are sliding over a surface
Rolling Friction - friction that acts on objects when they are rolling over a surface
Fluid Friction - friction that acts on objects that are moving through a fluid
Explanation:
Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book
, car parked on a hill.
Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.
Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.
Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon, sucking water through a straw, submarine moving through water.
Answer:The sun, earth, and moon are held together by gravity, and they interact in lots of ways.
- The tides are another interaction in the sun-earth-moon system. The tides happen because the moon and sun pull on the oceans, causing them to rise and fall each day. The moon has a bigger effect than the sun because it is closer.
Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive
Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:
Replacing the dat we obtain F=82 nN.
The force is repulsive because the points charged have the same sign.
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
