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viva [34]
3 years ago
11

Is loudness affected by change in frequency? Explain I will mark you the brainliest

Physics
1 answer:
zmey [24]3 years ago
4 0

Answer:

No

Explanation:

To increase or decrease loudness, you have to change the amplitude

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A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam
Allushta [10]
For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹ 
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
4 0
3 years ago
Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

Answer:

\Delta U = 1640 J

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

W = 560 J

so we have

\Delta U = Q - W

\Delta U = 2200 - 560

\Delta U = 1640 J

6 0
3 years ago
What happens to the magnitude of the gravitational force as the distance between two bodies increase?
Anna [14]

Answer:

The magnitude of the force will decrease

Explanation:

The gravitational force is one of the four fundamental forces of nature. It is an attractive force exerted between every object having mass.

Its magnitude is given by the equation:

F=\frac{Gm_1 m_2}{r^2}

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the separation between the objects

As we see from the equation, the magnitude of the gravitational force is inversely proportional to the square of the distance between the objects:

F\propto \frac{1}{r^2}

Therefore, this means that as the distance between two bodies increases, the gravitational force will decrease.

7 0
2 years ago
Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If
Anarel [89]
I think the answer would be Car C.
7 0
3 years ago
Read 2 more answers
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
2 years ago
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