All categories

2 years ago

What are the equations of angular acceleration, angular velocity and theta?.

Physics

1 answer:

soldier1979 [14.2K]2 years ago

3 0

Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t

You might be interested in

The stopwatch used by a student to measure velocity of a pulse in a slinky was of least count 0.1 second. He stops the stopwatch

sweet-ann [11.9K]

Least count of the pulse stopwatch is given by

$\Delta t = 0.1 s$

this means each division of the stopwatch will measure 0.1 s of time

After 3 journeys from one end to other we can see that total time that is measured here is shown by the clock as 52nd division

So here total time is given as

Time = (Number of division) (Least count)

now we will have

$T = 52 \times 0.1s$

$T = 5.2 s$

4 0

3 years ago

A sinusoidal wave traveling on a string has a period of 0.20 s, a wavelength of 32 cm, and an amplitude of 3 cm. The speed of th

Finger [1]

Answer:

$v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}$

Explanation:

If we have a periodic wave we need to satisfy the following basic relationship:

$v = \lambda f$

From the last formula we see that the velocity is proportional fo the frequency.

For this case we have the following info given by the problem:

$T= 0.2 s, \lambda =32 cm* \frac{1m}{100cm} =0.32 m, A= 3cm*\frac{1m}{100 cm}=0.03 m$

We know that the frequency is the reciprocal of the period so we have this formula:

$f = \frac{1}{T}$

And if we replace we got:

$f =\frac{1}{0.2 s}= 5Hz$

Now since we have the value for the wavelength we can find the velocity like this:

$v = 0.32 m * 5Hz = 1.6 \frac{m}{s}$

And if we convert this into cm/s we got:

$v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}$

6 0

2 years ago

What is the average speed of a boy who jogs 250 meters in 110 seconds

Ede4ka [16]

2.27 mps repeating.

This is the last question ill ever answer here. Thanks for being the last.

8 0

3 years ago

Read 2 more answers

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a

timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack ( $v$ ), measured in meters per second, is equal to the product of the angular speed of the wheel ( $\omega$ ), measured in radians per second, and the distance of the tack respect to the rotation axis ( $R$ ), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second: