Answer:
T0 =0, T1=0
Step-by-step explanation:
Please kindly go through the attached files for a detailed explanation.
I arranged them from the first one to the last one.
8.95 x 2 = 17.90
23.65 - 17.90 = 5.75
The answer is £5.75
I don't see the question . What is the question
Answer:
The time interval when
is at 
The distance is 106.109 m
Step-by-step explanation:
The velocity of the second particle Q moving along the x-axis is :

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.
We are also to that :
between 
The schematic free body graphical representation of the above illustration was attached in the file below and the point when
is at 4 is obtained in the parabolic curve.
So,
is at 
Taking the integral of the time interval in order to determine the distance; we have:
distance = 
= 
= By using the Scientific calculator notation;
distance = 106.109 m
Answer: 177.4999849 , or 56.5pi
Step-by-step explanation:
You simply divide the diameter by 2 to get the radius (28.25) and then plug into the circumference equation, 2*pi*r = Circumfrence.