Answer:
I. Friction force exerted on the body is less than 100N
Explanation:
For a body to be static, the moving force must be equal to the frictional force. Since the frictional force is a force of opposition. It tends to oppose the moving force acting on an object.
Hence if the moving force is greater than the force of friction, the Force of fiction will not be able to overcome the moving hence the body will tend to move.
Therefore, for a body to move, Fm > Ff or Ff < Ff
Fm is the moving force
Ff is the force of friction
Given
Fm = 100N
For the 100N body to move the frictional force must be less than 100N
Dependent on what you are measuring and what took you are using. Please be more specific.
The answer is 24 J
F K =.25*8 N
= 2N
F = f k = 2 N
Since a = 0
W = f * s
2 N * 12 m = 24 J
The coefficient of friction is a ratio used to quantify the friction force among two gadgets when it comes to the everyday pressure this is keeping them collectively. The coefficient of friction is critical attention at some stage in material selection and floor requirement determination.
For instance, ice on steel has a low coefficient of friction – the 2 materials slide past each different without problems – whilst rubber on the pavement has an excessive coefficient of friction – the substances no longer slide past each other without difficulty.
The coefficient of friction is dimensionless and it does not have any unit. it is a scalar, meaning the direction of the force does not have an effect on the physical quantity. The coefficient of friction depends on the gadgets that are causing friction.
Learn more about the coefficient of friction here brainly.com/question/20241845
#SPJ4
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
