Answer:
maximum amplitude = 0.13 m
Explanation:
Given that
Time period T= 0.74 s
acceleration of gravity g= 10 m/s²
We know that time period of simple harmonic motion given as
ω = 8.48 rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
10= 8.48² A
A=0.13 m
maximum amplitude = 0.13 m
Answer:
Explanation:
It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.
Let in equilibrium , tension in rope be T
For balancing Joe
T = M g
For balancing Simon
friction + T = mgsinθ
μmgcosθ+T = mgsinθ
μmgcosθ+Mg = mgsinθ
M = (msinθ - μmcosθ)
M = m(sinθ - μcosθ)
Answer:
Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.
Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.
Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).
Explanation:
Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.
Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.
Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).
This is the answer
Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.
The centripetal acceleration of any object moving in a circle is
(speed-squared) / (radius of the circle) .
Notice that we won't need to use the mass of the train.
We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.
Speed = (distance traveled) / (time to travel the distance).
Distance = 10 laps of the track. Well how far is that ? ? ?
1 lap = circumference of the track = (2π) x (radius) = 2.4π meters
10 laps = 24π meters.
Time = 1 minute 20 seconds = 80 seconds
The trains speed is (distance) / (time)
= (24π meters) / (80 seconds)
= 0.3 π meters/second .
NOW ... finally, we're ready to find the centripetal acceleration.
<span> (speed)² / (radius)
= (0.3π m/s)² / (1.2 meters)
= (0.09π m²/s²) / (1.2 meters)
= (0.09π / 1.2) m/s²
= 0.236 m/s² . (rounded)
If there's another part of the problem that wants you to find
the centripetal FORCE ...
Well, Force = (mass) · (acceleration) .
We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>
