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Ann [662]
2 years ago
11

A cube has a density of 1900 kg/m 3 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experime

nter in the laboratory as the cube moves through the laboratory at 92.0 %% of the speed of light in a direction perpendicular to one of its faces
Physics
1 answer:
Deffense [45]2 years ago
3 0

Answer:

4847.94844926 kg/m³

Explanation:

\rho' = Actual density of cube = 1900 kg/m³

\rho = Density change due to motion

v = Velocity of cube = 0.92c

c = Speed of light = 3\times 10^8\ m/s

Relativistic density is given by

\rho=\dfrac{\rho'}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-\dfrac{0.92^2c^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-0.92^2}}\\\Rightarrow \rho=4847.94844926\ kg/m^3

The cube's density as measured by an experimenter in the laboratory is 4847.94844926 kg/m³

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As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
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Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

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If in a vernier callipers 10 VSD coincides with 8 MSD then the least count of varnier calliper is
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A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

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= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

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= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

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