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neonofarm [45]
3 years ago
7

When 560,000 is written in scientific notation, what is the power of 10?

Physics
2 answers:
Rudiy273 years ago
6 0

Answer: 10^5

Explanation:

tensa zangetsu [6.8K]3 years ago
3 0
5.6•10^5 so it’s to the power of positive 5
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One mole of an ideal gas expands isothermally from 0.01 m3 to 0.05 m3. The final pressure of gas is 1.2 atm. [15 Marks] determin
erik [133]

Answer:

Initial pressure = 6 atm. Work = 0.144 J

Explanation:

You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J

4 0
3 years ago
An object to charge 2.00 c is a pretty from a second object with the same charge by a distance of 1.50 m what is the electric fo
Orlov [11]
F = q₁q₂C / r²

F force
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5 0
3 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
2 years ago
Please help this is really important thanks
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6 0
3 years ago
Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi
Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

I = I_0 cos^2\theta

Where,

I_0 = Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

\theta= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

I_1= \frac{I_0}{2}

In the case of the second polarizer the angle is directly 60 degrees therefore

I_2 = I_1 cos^2\theta

I_2 = (\frac{I_0}{2} ) cos^2(60)

I_2 = 0.125I_0

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

\theta_3 = 90-60 = 30

Then,

I_3 = I_2 cos^2\theta_3

I_3 = 0.125I_0 cos^2 (30)

I_3 = 0.09375I_0

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

5 0
2 years ago
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