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3 years ago

3. Which of the following is NOT a function of the liver?

Physics

2 answers:

Inga [223]3 years ago

8 0

Answer:

C) to produce white blood cells

Explanation:

Your liver does all of the others things except produce white blood cells

If it helps consider making brianliest (it helps alot)

mash [69]3 years ago

7 0

I think it’s C, that persons answer is good

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A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?

melisa1 [442]

The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =200

K is the spring constant= 2500 N/m

x is the length by which spring got stretched =?

The stretch of the spring is found as;

$\rm F=kx \\\\ x = \frac{F}{k} \\\\\ x= \frac{200}{2500} \\\\ x=0.08 \ m$

Hence the length by which the spring got stretched will be 0.08 m.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ4

8 0

2 years ago

Read 2 more answers

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

3 years ago

Who was the first scientist to propose that an object could emit only certain amounts of energy?

Rzqust [24]

It was Niels Bohr who proposed it

3 0

3 years ago

A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

3 0

3 years ago

Knowledge and skills learned through socialization are an example of

ziro4ka [17]

Answer:

I think no.2 the answer

Because socialization and social resources are both for me

3 0

3 years ago