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2 years ago

How would you characterize William's relationship with his father? In what ways did his father help shape William's outlook?

Physics

1 answer:

Aleksandr [31]2 years ago

3 0

William's father shaped William's outlook on life when he was younger due to his bravery.

<h3>What was the relationship between William and his father?</h3>

The impact of significant others in the life of a person can not be over emphasized. These significant others often serve as role models and help to develop the individuality and personality of people.

The life of William was shaped quite well by his father because the both of them had a cordial relationship with each other. William's father shaped William's outlook on life when he was younger due to his bravery.

Learn more about William and his father:brainly.com/question/4066623

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A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$

The potential energy reached by the ball at its rebound is

$U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J$

Thus the speed after the collision is

$\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}$

$v'=13.17\ m/s$

The change of momentum is

$\Delta P=0.0616\cdot (13.17+14)$

$\Delta P=1.67 \ kg.m/s$

(b) If the putty sticks to the floor, then v'=0

$\Delta P=0.0616\cdot (0+14)$

$\Delta P=0.86\ m/s$

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