-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the
1 answer:
Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
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Answer:
Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:
Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:
Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:
Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So and
(Since
). Then, we get:
In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
The word/prefix "
" means 
Good luck on your assignment and enjoy your day!
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Good luck on your assignment and enjoy your day!
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