1) Classify each of the following as either a solid, liquid or gas at room temperature.

Can someone help me please? I’m so confused

Svetradugi [14.3K]

Answer:

what do you need help with? Like which type of print each image is?

Explanation:

5 0

3 years ago

A guitar string is supposed to have a fundamental frequency 256 Hz. It currently has a fundamental frequency 248 Hz when the str

umka2103 [35]

Answer:

The tension to bring the guitar string into tune is 372.95 Hz.

Explanation:

Given;

current frequency, f₁ = 248 Hz

current tension, T₁ = 350 N

fundamental frequency, f₂ = 256

The tension on the string to bring the guitar string into tune is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\f\lambda = \sqrt{\frac{T}{\mu} } \\\\f^2\lambda^2 = \frac{T}{\mu} \\\\f^2 = \frac{T}{\mu \lambda^2}\\\\let \ {\mu \lambda^2} = k\\\\f^2 =\frac{T}{k} \\\\k = \frac{T}{f^2} \\\\\frac{T_1}{f_1^2} = \frac{T_2}{f_2^2}\\\\T_2 = \frac{T_1 f_2^2}{f_1^2} \\\\T_2 = \frac{350 \times 256^2}{248^2} \\\\T_2 = 372.95 \ Hz$

Therefore, the tension to bring the guitar string into tune is 372.95 Hz.

3 0

3 years ago

What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is

QveST [7]

Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
</span> $F=qE$
<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span> $E=k_e \frac{Q}{r^2}$
with $k_e = 8.99\cdot 10^9 Nm^2/C^2$ , $Q=+25 nC=25 \cdot 10^{-9}C$ is the charge on the bead. We want to calculate the field at $r=4.0 cm=0.04 m$ :
$E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m$
The proton has a charge of $q=1.6\cdot 10^{-19}C$ , therefore the force exerted on it is
$F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N$

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, $m=1.67\cdot 10^{-27} kg$ , we have
$F=ma$
$a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2$

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
$a=-1.35 \cdot 10^{13} m/s^2$

6 0

4 years ago

At higher elevations, the boiling point of water decreases, due to the decrease in atmospheric pressure. As a result, what

Effectus [21]

Answer:

The correct answer is option b. At higher elevations, it would take longer to hard boil an egg, because there is a lower boiling.

Explanation:

When we boil water at a very high temperature, it turns out that the water boils much faster. This is because the higher we are, <u>there will be less oxygen and less pressure in the atmosphere</u>. This implies that the water vapor becomes steam much faster than as it happens at low temperatures, so less heat will be needed for the water to boil.

That is why <u>if we want to cook an egg, we will have to lower the heat</u> of the kitchen when we put it on the fire, in order to avoid boiling water very quickly. Lowering the heat, <u>then cooking the egg will take much longer. </u>

Given this data, the correct answer is option B, At higher elevations, it would take longer to hard boil an egg, because there is a lower boiling.

8 0

4 years ago

Three resistors are connected in parallel. If placed in a circuit with a 12-volt power supply. Determine the equivalent resistan

Ivan

(a) The equivalent resistance of three parallel resistors is (R₁R₂R₃)/(R₁R₂ + R₁R₃ + R₂R₃).

(b) The total circuit current is 12/R.eq.

(c) The voltage drop across and current in each resistor is IR₁, IR₂ and IR₃.

<h3>Equivalent resistance of three parallel resistors</h3>

The equivalent resistance of three parallel resistors is calculated as follows;

Let the first resistor = R₁
Let the second resistor = R₂
Let the third resistor = R₃

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_2}\\\\R_{eq} = \frac{R_1R_2R_3}{R_1R_2 + R_2R_3 + R_1R_3}$

<h3>Total Circuit Current </h3>

The total circuit current is calculated as follows;

V = IR

$I= \frac{V}{R_{eq}} = \frac{12 \ V}{R_{eq}}$

<h3>Voltage drop in each resistor</h3>

Voltage drop in resistor 1 = IR₁

Voltage drop in resistor 2 = IR₂

Voltage drop in resistor 3 = IR₃

Learn more about parallel circuit here: brainly.com/question/80537

#SPJ1

3 0

2 years ago