AP PHYSICS please answer

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Answer 1

Answer: I think it’s d not sure

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IWhat is your mass in kilograms of you weigh 120 pounds?

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One pound is (around, actually it's a little more, but you have to round it down to make the number practically usable) 0.45 kilograms, so 120 pounts is 120* 0.45, which is around 54kg.

You use a 1200-watt hair dryer for 10 minutes each day. a. How many minutes do you use the hair dryer in a month? (Assume there are 30 days in the month.) b. How many hours do you use the hair dryer in a month? c. What is the power of the hair dryer in kilowatts? d. How many kilowatt-hours of electricity does the hair dryer use in a month? e. If your town charges $0.15/kWh, what is the cost to use the hair dryer for a month?

Answers

Data;

Power = 1200W
Time/day = 10 mins

Power

The is the rate of energy used to time. It is the measure of expedience of energy.

a). How many minutes is the dryer used in a month.

Assuming there 30 days in a month;

$1 day = 10 mins\n30 days = x mins\nx = 30 * 10 \nx = 300 mins$

This shows that the dryer is used 300 minutes in a month.

b). How many hours is the dryer used in a month.

To solve this problem, we simply need to convert the value of (a) from minutes to hours.

$1hr = 60 mins\ny hr = 300 mins\ny = (300)/(60) \ny = 5 hours$

The dryer was used for a total of 5 hours in a month.

c). The power of the dryer in kilowatt.

To convert the power in watt to kilowatt, we divide by 1000.

$P = 1200W\nP = (1200)/(1000) \nP = 1.2kw$

The power in kilowatt is 1.2kw

d). How many kilowatt/hr of electricity used in a month.

$1.2 * 5 = 6kw/h$

The dryer consumes 6kw/h in a month.

e). If the town charges $0.15/kwh, the cost in a month

$\$0.15 = 1hr\nx= 6hr\nx = 6 * 0.15\nx = 0.9$

The cost of the dryer in a month is $0.9.

Learn more on power here;

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300 minuts
5 hours
1.2 killowatt
6 killowatt hour
6*0.15=0.9
$0.9

What is an example of a high amplitude sound, and an example of a low amplitude sound?

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Rock concerts and whispers are examples of a high-amplitude sound and a low-amplitude sound.

The largest displacement of sound wave constituents from their resting positions is referred to as amplitude. It stands for the loudness or intensity of a sound, to put it simply. Here are some illustrations of both high and low-amplitude sounds:

High Amplitude Sound: An illustration of a high amplitude sound is a rock concert with loudspeakers blaring songs at full intensity. The concert speakers produce sound waves with a tremendous amplitude, creating a powerful, strong sound that can be heard from a great distance.

Low Amplitude Sound: A low amplitude sound is something like the sound of a whisper. The sound created when someone whispers is calm and soft and not as loud as a rock concert, since the sound waves produced have a tiny amplitude.

In both cases, how loud or soft the sound is perceived by our ears depends on the amplitude of the sound waves. Low-amplitude sounds are soft and quiet, but high-amplitude sounds are strong and loud.

Hence, rock concerts and whispers are examples of a high-amplitude sound and a low-amplitude sound.

To learn more about Amplitude, here:

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#SPJ3

High amplitude is a sound of high loudness like that of traffic, DJ, and earthquake volcano etc.

Low amplitude is feeble sound like that of light breeze, or that of whispering

Radio waves travel at 3.00 · 108 m/s. Calculate the wavelength of a radio wave of frequency 900 kHz. (9.00 · 105 Hz.)_____ m

2.70 x 1014
3.70 x 10-15
3.33 x 102
3.00 x 10-3

Answers

v: velocity of wave
f: frequency
L: wavelenght

v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m

A single star in the process of forming starts by spinning slowly (while it is quite large and relatively cool.) As the star collapses under the pull of its own gravity, its size decreases. As a result, its rate of spinning:_____.

Answers

Answer:

Explanation:

According to the conservation of momentum, when the external torque is not applied on the system, then the angular momentum of the system is conserved.

I x ω = constant

$I_(1)\omega _(1)=I_(2)\omega _(2)$

where, I is the moment if inertia and ω is the angular velocity

As the star is collapsed, the radius decreases and hence the moment of inertia also decreases.

According to the law, the angular velocity increases, thus the rate of spinning increases.

A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistanceof 0.757 ohm at 20°C. Calculate the resistivity of the wire. [Show all work, including the equation and
substitution with units.]

Answers

The resistivity of the plantinum wire is 1.06 × 10⁻⁷Ωm

Calculating Resistivity

From the question,

We are to determine the resistivity of the wire

Resistivity can be calculated from the formula

$\rho = (RA)/(l)$

Where $\rho$ is the resistivity of the wire

R is the resistance

A is the cross-sectional area

and $l$ is the length of the wire

From the given information,

R = 0.757 ohm (Ω)

A = 3.50 × 10⁻⁶ m²

$l$ = 25.0 m

Putting the parameters into the formula, we get

$\rho = (0.757 * 3.50 * 10^(-6) )/(25)$

$\rho = 0.10598 * 10^(-6)$

$\rho = 1.0598 * 10^(-7)$

$\rho \approx 1.06 * 10^(-7) \ \Omega m$

Hence, the resistivity of the plantinum wire is 1.06 × 10⁻⁷Ωm

Learn more on Calculating Resistivity here: brainly.com/question/5595373

R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m

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