Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>
Solution:
As we know that displacement is calculated in centimeters and the unit of time is second.
The average velocity for the first interval [1,2] is given
Δs / Δt = s (t2) - s (t) / t2 - t1
Δs / Δt = 2sin2 π + 3cos 2 π - ( 2sin π + 3cos π ) / 2 - 1
Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1
Δs / Δt = 6 cm/s
Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s
If you need to learn more about displacement click here:
brainly.com/question/28370322
#SPJ4
The complete question is:
The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1
Answer:
since -6 lasted for 5 seconds, multiplying both would result in -30
3 lasted for 10 seconds, so multiplying both would give +30
average = ( 30 + (-30) ) / 2
30 -30 is already equal to zero, so the answer should be 0
I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
Answer:
Distance covered by B is 4 times distance covered by A
Explanation:
For an object in free fall starting from rest, the distance covered by the object in a time t is
where
s is the distance covered
g is the acceleration due to gravity
t is the time elapsed
In this problem:
- Object A falls through a distance 
during a time t, so the distance covered by object A is
- Object B falls through a distance 
during a time 2t, so the distance covered by object B is
So, the distance covered by object B is 4 times the distance covered by object A.
