You are given the speed of a wave and it's period. What kind of information can you also find out about this wave?

Which would be a common-sense practice in a lab environment?

algol13

Using safe research practices would be a very good common-sense practice in a lab environment.

6 0

2 years ago

Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri

Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

(c) 0.716 s

Explanation:

Gravity would be exerting on the 2 masses

$G_1 = Mg = 15.3*9.81 = 150.093N$

$G_2 = mg = 8.29*9.81= 81.32N$

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

$F = G_1 - G_2 = 68.77N$

This force would generate torque on the solid pulley

$T = FR = 68.77 * 0.0075 = 0.5158 Nm$

We can also calculate the pulley moments of inertia, with it being solid

$I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2$

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

$\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2$

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

$\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad$

(c)The time it takes to get to this distance is

$\theta = \frac{\alpha t^2}{2}$

$t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513$

$t = \sqrt{0.513} = 0.716s$

(b)The final angular speed of the disk is

$\omega = \alpha t = 1300.58*0.716 = 931 rad/s$

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

$v = \omega R = 931*0.0075 \approx 7m/s$

5 0

2 years ago

A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both e

xxMikexx [17]

Answer:

the wave represents the second harmonic.

Explanation:

Given;

length of the cord, L = 64 cm

The first harmonic of a cord fixed at both ends is given as;

$f_o = \frac{V}{2L}$

The wavelength of a standing wave with two antinodes is calculated as follows;

L = N---> A -----> N + N ----> A -----> N

Where;

N is node

A is antinode

L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2

L = λ

The harmonic is calculated as;

$f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic$

Therefore, the wave represents the second harmonic.

L = λ

5 0

2 years ago

Why is it impossible to create a perpetual motion machine?

Tatiana [17]

Answer:

It is impossible to create a perpetual motion machine because some of the energy will always be lost in the conversion and therefore it will eventually stop.

Correct Answer : Option B

Explanation:

The perpetual motion machine is impossible machine as it is hypothetical working machine which would be in motion for an indefinite time in continuity of the motion. The continuity of motion for an indefinite time would mean that the working principle of the machine would never allow the dissipation of energy from the machine and all the machine would ultimately reserve all the energy and be converting it into forms without any loss.

This working principle is violation of first and second laws of thermodynamics and hence a machine will eventually lose some of its energy in every conversion of the working cycle and hence there will be a time where the machine would be stopped, and hence a perpetual motion machine cannot be made.

8 0

2 years ago

A stationary boat bobs up and down with a period of 2.4 s when it encounters the waves from a moving boat.

lapo4ka [179]

It seems like the question is asking for the frequency.
Given:
Time period (T) = 2.4 sec

Frequency (f) =?

We know that the formula for frequency is:

Frequency (f) = 1/time period (T)

= 1 / 2.4 s

= 0.42 Hz. is the frequency for this problem.

8 0

2 years ago