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wlad13 [49]
3 years ago
5

When comparing a star’s absolute magnitude to its apparent magnitude, which is usually the case? Select all that apply.

Physics
1 answer:
kirza4 [7]3 years ago
3 0

All four choices are true statements !

(I've never seen that before in a Brainly question.)

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In being served, a tennis ball is accelerated from rest to a speed of 42.1 m/s. The average power generated during the serve is
hoa [83]

Answer:

Force F = 69.35 N

Explanation:

given data

Ball Initial speed u = 0

Ball Final speed v = 42.1 m/s

average power generate = 2920 W

solution

Power generate is express as

P= \frac{W}{t}   ..............1

here W is work done and t is time

and work w = F × d

so

P= \frac{Fd}{t}  

and we know speed v = \frac{d}{t}  

so here

Power P = F × v

put here value and we get force

Force F = \frac{2920}{42.1}

Force F = 69.35 N

8 0
4 years ago
Is O2 classified as a compound?
dybincka [34]

Answer: NO.

Explanation: Oxygen is not a compound. It has only one element in it.

3 0
3 years ago
) 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
Physics!!! please help D:
Ugo [173]
Answer: A

Explanation: isotopes of the same thing element have the same number of protons in the nucleus but differ in the number of neutrons.
6 0
3 years ago
An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac
elena-s [515]

First of all, we need to convert the angular speed from rev/min into rev/s:

\omega_f=300 rev/min=5 rev/s

The angular acceleration is the variation of angular speed divided by the time:

\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev

so, 5 revolutions.

3 0
3 years ago
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