A lab cart is loaded with different masses and moved at various constant velocities? the anser should be
1.0m/s → 4kg
Angular acceleration = (change in angular speed) / (time for the change)
change in angular speed = (zero - 2,600 RPM) = -2,600 RPM
time for the change = 10 sec
Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec
(-260 rev/min-sec) x (1 min / 60 sec) = <em>-(4 1/3) rev / sec²</em>
Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .
The average speed is 1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.
Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = <em>(216 2/3) revs</em>
Answer:
We know that for a pendulum of length L, the period (time for a complete swing) is defined as:
T = 2*pi*√(L/g)
where:
pi = 3.14
L = length of the pendulum
g = gravitational acceleration = 9.8 m/s^2
Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.
Then the period is independent of:
The mass of the child
The initial angle
Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.
Answer:
Explanation:
firstly we have to find the frequency and then the angular frequency.
The frequency will be:
The angular frequency is:
ω
ω
ω
Now we can apply simple pendulum relationship
ω
make g the subject of the formula
