What is the insertion for the brachialis

When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room. Which sentence explains t

Katyanochek1 [597]

When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room because gases do not have a definite volume.

Hope this helps!

5 0

3 years ago

What do all waves carry

MAVERICK [17]

Answer:

the waves carry energy

Explanation:

this is the way you can write it in short manner

hope it helps you.

MARK me as BRAINLIEST pls

5 0

2 years ago

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

so, the difference between any two harmonics tube is equal to:

$f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 4699 Hz\\f_{n+1}=4953 Hz$

So, according to (1), the fundamental frequency is equal to half of this difference:

$f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz$

Now we know that one of the harmonics is $f_n=4191 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz$

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

Since we know $f_n = 4445 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17$

7 0

3 years ago

580 nm light shines on a double slit

Helga [31]

Answer:

0.532

Explanation:

Your equation to find the second bright interference maximum is gonna be this: d sin (Θ) = m λ

First, find your variables.

λ = 580 · 10^-9

d = 0.000125

m = 2

Next, fill in the equation.

d sin (θ) = m λ

(0.000125) sin (θ) = (2) (580·10^-9)

Then isolate your variable.

θ = arcsin ( (2)(580·10^-9) / (0.000125) )

Run your equation and you will end up with 0.53171246 , which rounds to 0.532.

The main thing you have to watch out for is make sure you are calculating for the bright interference and not the dark interference, as well as checking you're calculating for the maximum, not the minimum.

I hope this helps :D

5 0

3 years ago

6) If Bam was hit by the hand with a force of 1200N. If he has a mass of 81kg, what acceleration did

bazaltina [42]

Answer:

14.81 m/s^2

Explanation:

f=m×a

1200=81×a

a=1200/81

a=14.81

3 0

3 years ago