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avanturin [10]
3 years ago
6

A bell is rung. What best describes the density of air around the bell? The air density does not change. The air density increas

es and then returns to normal levels. The air density decreases and then returns to normal levels. The air density increases and decreases repeatedly before returning to normal.
Physics
2 answers:
dezoksy [38]3 years ago
7 0

I think the closest possible answer to this question is The air density increases and decreases repeatedly before returning to normal.Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
Yakvenalex [24]3 years ago
4 0

The correct answer to the question is : D). The air density increases and decreases repeatedly before returning to normal

EXPLANATION:

Before coming into any conclusion, first we have to understand sound wave.

A sound wave is a longitudinal wave in which the vibration of particles are parallel to the direction of wave propagation. It moves in the form of compression and rarefaction.

Compression is the high pressure region where particles are closely aggregated. Rarefaction is the region of low pressure where particles of the medium are not so close to each other just like compression.

Hence, the correct statement  will be that the air density increases and decreases repeatedly before returning to normal .

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A student on an amusement park ride moves in a circular path with a radius of 3.5 meters once every 8.9 seconds. What is the ave
Ulleksa [173]
Distance traveled by him = circumference of that circular path = 2πr = 2π(3.5)
= 7π = 7×3.14 = 21.98 m
time = 8.9 s  [ Given ]

Now, Average speed = distance / time
s = 21.98 / 8.9
s = 2.46 m/s

Hope this helps!
7 0
3 years ago
Four pairs of objects have the masses as described below, along with the distances between
lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

The gravitational force is defined as

F=G\frac{m_{1} m_{2} }{r^{2} }

By definition, the gravitational force depends directly on the product of the masses and indirectly on the distance between the masses, which means the further they are, the less gravitational force would be. And, the greater the masses, the greater the gravitational force.

Among the options, the pair that would have the greatest gravitational force is  Mass of 1 Kg and 2 Kg, with 1 meter between them.

Notice that the last choice includes the same masses but with a greater distance between them, that means it would be a weaker graviational force.

Therefore, the right answer is the second choice.

7 0
3 years ago
Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0
3 years ago
Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini
natka813 [3]
E=energy=5.09x10^5J = 509KJ 
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
5 0
3 years ago
An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna
Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

I=\frac{V}{R}

Where R is the equivalent resistance of the resistors in series

R=0.0510+0.0090=0.0600[ohm]

I=\frac{12.0}{0.0600}=200A

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V

The power dissipated supplied to the motor is given by:

P=I^2*R_m\\P=(200)^2*0.0510=2.04kW

now solving adding a 0.0900 ohm resistor:

I=\frac{12.0}{0.15}=80A

V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V

P=I^2*R_m\\P=(200)^2*0.0510=0.326kW

5 0
3 years ago
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