Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
The answer is %pearlite = 0.06%
Explanation:
according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:
%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%
The percentage of cementite is equal to:
%cementite = (1.88-0)/(2-0) = 0.94%
The person's horizontal position is given by

and the time it takes for him to travel 56.6 m is

so your first computed time is the correct one.
The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity
at time
is

which tells us that he would reach the ground at about
. In this time, he would have traveled

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity
would have been a bit smaller than
.
Answer:
Force, F = 77 N
Explanation:
A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.
We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :
F = m × a
Where
m = mass = 55 kg
a = acceleration = 1.4 m/s²

F = 77 N
So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.
It's commonly referred to as an electric current.