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3 years ago

6

Please help I don't understand. A teacher gives students four liquids commonly found in the kitchen - vinegar, apple juice, dish

detergent, and milk - and pH indicator strips, which measure acidity. The teacher asks the students to put the liquids in order from most acidic to most basic.
What type of investigation could the students conduct to determine the correct order for the liquids? 1. descriptive 2.Comparitive 3. experimental
The type of liquid is 1. independent variable 2. dependant variable 3. control group
The pH is 1. independent variable 2. dependant variable 3. control group

1 answer:

hram777 [196]3 years ago

3 0

Answer:

Its Experimental

Explanation:

the others don't make sense

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A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi

OverLord2011 [107]

Answer:

$P=214.7187\,W$

Explanation:

Given that:

Diameter of the solenoid, $D=10\,cm=0.1\,m$

length of the solenoid, $L=90\,cm=0.9\,m$

diameter of the wire, $d=0.1\,cm=10^{-3}\,m$

magnetic field at the center of the solenoid, $B=7.4\times 10^{-3}\,T$

<u>Now we need the no. of turns incorporated in the length of 90 cm:</u>

$N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}$

$N=\frac{L}{d}$

$N=\frac{0.9}{10^{-3}}$

$N=900\,\,turns$

For solenoids we have:

$B=\mu.n.I$ ...............................(1)

where:

$\mu=$ permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

$n=\frac{900}{0.9}$

$n=1000\,turns\,.\,m^{-1}$

Now putting the respective values in the eq. (1)

$7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I$

$I=5.8887\,A$

For copper we have resistivity:

$\rho=1.72\times 10^{-8}\, \Omega.m$

We know that resistance is given by:

$R=\rho.\frac{l}{a}$ .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

<u>Now we need the length (l) of the wire:</u>

Circumference of the solenoid,

$C=\pi.D$

$C=0.1\pi\,m$

$\therefore l=C\times N$

$l=90\pi\,m$

&

<u>Cross-sectional area of wire:</u>

$a=\pi.\frac{d^2}{4}$

$a=\pi. \frac{(10^{-3})^2}{4}\,m^2$

<u>Resistance from eq. (2):</u>

$R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}$

$R=6.192 \,\Omega$

For power we have:

$P=I^2.R$

$P=5.8887^2 \times 6.192$

$P=214.7187\,W$

6 0

3 years ago

Two musicians are comparing their trombones. The first produces a tone that is known to be 438 Hz. When the two trombones play t

galben [10]

Answer:

It is producing either a 435-Hz sound or a 441-Hz sound.

Explanation:

When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:

$f_B = |f_1 -f_2|$ (1)

In this problem, we know that:

- The frequency of the first trombone is $f_1 = 438 Hz$

- 6 beats are heard every 2 seconds, so the beat frequency is

$f_B=\frac{6}{2 s}=3 Hz$

If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:

$f_2 = f_1 - f_B = 438 Hz - 3 Hz = 435 Hz\\f_2 = f_1+f_B = 438 Hz+3 Hz=441 Hz$

7 0

3 years ago

What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep

kari74 [83]

Answer:

708.3 Hz

Explanation:

For an open-air column, like the empty can, the fundamental frequency is given by

$f_1 = \frac{v}{2L}$

where

v = 340 m/s is the speed of sound

L is the length of the column

In this problem, the length of the bottle is

L = 24 cm = 0.24 m

Therefore, the fundamental frequency is

$f_1 = \frac{340 m/s}{2(0.24 m)}=708.3 Hz$

8 0

3 years ago

What is the objects acceleration

alekssr [168]

Answer:

Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

5 0

2 years ago

Read 2 more answers

The part of the electromagnetic spectrum usually used for communication is

noname [10]

<h2>Answer:</h2>

The part of the electromagnetic spectrum usually used for communication is of radio waves, infrared waves and microwaves.

But most common waves used in communication are radio waves.

<h3>Explanation:</h3>

Radio waves are used to transmit television and radio programmes.
While microwaves are used for mobile phones and Wi-Fi.
Radio waves are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light.
Radio waves have frequencies as high as 300 gigahertz (GHz) to as low as 30 hertz (Hz).

5 0

3 years ago