Answer: (a) t1 = omega1/alpha
(b) theta1 = 1/2 * alpha*theta1^2
(c) t2 = omega2/5*alpha
Explanation: see attachment
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
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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:
0 J
Explanation:
= Work done on the satellite in circular orbit about earth by earth
= Force on gravity on satellite by earth
= displacement of the satellite
= Angle between the force on gravity and displacement = 90
We know that, Work done is given as
Answer:
work relationship built on trust.
Explanation:
Align on a set of communication best practices understood by all members of the team. Create a work environment built on trust and empathy. Manage healthy conflict through standards, processes, and real-time feedback. Develop a support system through patience and discipline.
Answer:
Explanation:
Initial height from the ground = .41 m
Final height = 1m
Height by which Kelli was raised ( h )= .59 m
When she passes through the lowest point , she loses P E
= mgh
= 440 x .59
= 259.6 J
kinetic energy possessed by her
= 1/2 mv²
= .5 x (440/9.8) x 2²
= 89.8 J
Difference of energy is lost due to work by air friction
work done by friction = 89.8 - 259.6
= - 169.8 J
