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valentina_108 [34]

3 years ago

5

Place these bodies of our solar system in the proper order of formation

2 answers:

Flauer [41]3 years ago

8 0

Answer:

solar nebula, the sun, planetesimals, inner planets, and last but not least outer planets

meriva3 years ago

5 0

Answer:

Solar nebula, The sun, planetesimals , inner planets, outer planets

Explanation:

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) 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

PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!

Liula [17]

The answer is Trend Line.

5 0

3 years ago

Read 2 more answers

Which bone would a forensic anthropologist analyze to identify a victim as male or female?

noname [10]

the femmus is the answer

6 0

3 years ago

A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string

Darya [45]

Answer:

a

The speed of wave is $v_1 = 129.1 \ m/s$

b

The new speed of the two waves is $v = 129.1 \ m/s$

Explanation:

From the question we are told that

The mass of the string is $m = 60 \ g = 60 *10^{-3} \ kg$

The length is $l = 2.0 \ m$

The tension is $T = 500 \ N$

Now the velocity of the first wave is mathematically represented as

$v_1 = \sqrt{ \frac{T}{\mu} }$

Where $\mu$ is the linear density which is mathematically represented as

$\mu = \frac{m}{l}$

substituting values

$\mu = \frac{ 60 *10^{-3}}{2.0 }$

$\mu = 0.03\ kg/m$

So

$v_1 = \sqrt{ \frac{500}{0.03} }$

$v_1 = 129.1 \ m/s$

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

8 0

3 years ago

5. Which of the following is velocity? *

ivann1987 [24]

A. 20m/s because the unit for velocity is m/s

3 0

3 years ago