A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry
B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.
Answer: friction is used when 2 and 2 things mix and come together but normally friction slows things down but when buttering a pan it doesnt slow down
Answer:
THE RUBBER BALL
Explanation:
From the question we are told that
The mass of the rubber ball is 
The initial speed of the rubber ball is 
The final speed at which it bounces bank 
The mass of the clay ball is 
The initial speed of the clay ball is 
The final speed of the clay ball is 
Generally Impulse is mathematically represented as
where
is the change in the linear momentum so

For the rubber is


=> 
For the clay ball


=> 
So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball
The sound gets louder as it gets closer and when it passes is gets softer
Answer:
Yes is large enough
Explanation:
We need to apply the second Newton's Law to find the solution.
We know that,

And we know as well that

Replacing the aceleration value in the equation force we have,

Substituting our values we have,


The weight of the person is then,


<em>We can conclude that force on the ball is large to lift the ball</em>