Out of hypothesis, theories and laws “can help predict future events”

Noise-canceling headphones are an application of destructive interference. Each side of the headphones uses a microphone to pick

exis [7]

The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.

4 0

3 years ago

8. A driver starts his parked car and within 12 hours reaches a speed of 75km/h, as he travels

erastovalidia [21]

Answer:

56mph

Explanation:

7 0

2 years ago

A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

$m_c$ = The center of mass of the stick = $\frac{L}{2}-0.22=0.5-0.22=0.28\ m$

$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

4 0

3 years ago

A physical count of merchandise inventory on november 30

Zigmanuir [339]

It depends on the indirect taxes and the share market values

4 0

3 years ago

The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele

Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

$P=P_0e^{-\frac{Mg}{RT}h}$