A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain
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Answer:
I. Speed = 20m/s
II. Velocity = 20m/s due North.
Explanation:
<u>Given the following data;</u>
Distance = 40m
Time = 2secs
To find the speed;
Mathematically, speed is given by the formula;
Substituting into the equation, we have;
<em>Speed = 20m/s.</em>
In physics, we use the same formula for calculating speed and velocity. The only difference is that speed is a scalar quantity and as such has magnitude but no direction while velocity is a vector quantity and as such it has both magnitude and direction.
<em>Therefore, the velocity is 20m/s due North</em>.
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The sprinter’s average acceleration is 1.98 m/s²
The given parameters;
- initial velocity of the sprinter, u = 18 km/h
- final velocity of the sprinter, v = 27 km/h
- time of motion of the sprinter, t = 3.5 x 10⁻⁴ h
Convert the velocity of the sprinter to m/s;
The time of motion is seconds;
The sprinter’s average acceleration is calculated as follows;
Thus, the sprinter’s average acceleration is 1.98 m/s²
Learn more here:brainly.com/question/17280180
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find