What measures the radiation given off by earth

) 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

"My partner seems to be more in the mood at night time, but I'm more in the mood in the morning. Why might that be and is there

Marianna [84]

Answer:

I don't get it?

like yhu want us to rate it or?

Explanation:

5 0

3 years ago

Is this the actual answer??

Nastasia [14]

It probably is the actual answer.

3 0

3 years ago

A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free

Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R from equation (1)

T = I x a / R² .... (2)

mg - T = ma .... (3)

Substitute the value of T from equation (2) in equation (3)

$mg - \frac{Ia}{R^{2}}=ma$

$a=\frac{mg}{m+\frac{I}{R^{2}}}$

T is the acceleration in the system

Substitute the value of a in equation (2)

$T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}$

$T=\frac{I\times mg}{I+mR^{2}}$

This is the tension in the string.

4 0

3 years ago

The speed of light in a solid is 1.24 x 108 m/s. <br> Calculate the index of refraction

Dahasolnce [82]

Answer:

125

Explanation:

7 0

3 years ago