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2 years ago

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

Physics

1 answer:

ludmilkaskok [199]2 years ago

3 0

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$

$\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})$

$\log_{10}(b_{1}/b_{2}) = 3.44$

Thus,

$(b_{1}/b_{2}) = 2754.22$

It means star A is 2754.22 time brighter than Star B.

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2 years ago

The frequency of oscillation of a certain LC circuit is 299 kHz. At time t = 0, plate A of the capacitor has maximum positive ch

AnnZ [28]

Answer:

Explanation:

Given an LC circuit

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f = 299 kHz = 299,000 Hz

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t = 1 / f

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t = 0.00000334448 seconds

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t = 3.34 μs

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Where n = 1,2,3,4....

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where n = 1,2,3...

C. After every half of t,inductor have maximum magnetic field at time

t'' = ½ × t'

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3 years ago

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

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Internal energy, U, is equal to the work done or by the system, plus the heat of the system: ΔU=q+w
in the question they tell you the work done by the system, and the internal energy:
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substitute in: q=m∗Cp∗ΔT and solve for Cp.

-------------------------------------
remember that ΔT=Tf−Ti

so the equation, really, is: q=m∗Cp∗(Tf−Ti)
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185J=−346J+[m∗Cp∗(Tf−Ti)]

plug in the rest of your values and solve for Cp

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2 years ago

Through what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinho

Mila [183]

Answer:

0.15 mV

Explanation:

In order to exhibit wave nature, the de Broglie wavelength of the electron must be of the same size of the diameter of the pinhole, therefore:

$\lambda=0.10 \mu m = 1.0\cdot 10^{-7} m$

The de Broglie wavelength of an electron is

$\lambda = \frac{h}{mv}$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

v is the electron's speed

Therefore, the electron's speed must be

$v=\frac{h}{m\lambda}=\frac{6.63\cdot 10^{-34}}{(9.11\cdot 10^{-31})(1.0\cdot 10^{-7})}=7278 m/s$

When accelerated through a potential difference $\Delta V$ , the kinetic energy gained by the electron is equal to the change in electric potential energy, therefore

$e\Delta V = \frac{1}{2}mv^2$

where

$e=1.6\cdot 10^{-19}$ is the magnitude of the charge of the electron

So, we can find the potential difference needed:

$\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.5\cdot 10^{-4}V = 0.15 mV$

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3 years ago