A student clicks the mouse at time intervals of 1.60, 2.10, 2.40, 1.90 what is the standard deviation of this dataset.

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

7 0

3 years ago

A force of 2N will stretch a rubber band 0.02m. Assuming that Hooke's Law applies, answer the following: How far will a 1600N fo

denis-greek [22]

Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant

F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m

Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ

7 0

3 years ago

What is the current in a 160V circuit if the resistance is 2Ω? V= I= R=

Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

160 = 2I

2 2

I = 80A

4 0

3 years ago

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte

musickatia [10]

Answer:

Number of electrons, $n=5.62\times 10^{21}$

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, $I=\dfrac{q}{t}$

$I\times t=n\times e$

$n=\dfrac{It}{e}$

e is the charge of an electron

$n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}$

$n=5.62\times 10^{21}$

So, the number of electrons pass through the resistor is $5.62\times 10^{21}$ . Hence, this is the required solution.

6 0

3 years ago

Why does sodium chloride and magnesium chloride conduct electricity?

mihalych1998 [28]

They only conduct when they are in solution form, because then their ions become mobile, and the ions conduct electricity.

Hope this helps!

5 0

3 years ago