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xz_007 [3.2K]
3 years ago
11

The ratio of brightness of two stars can be determined by a constant of 2.512 raised to the power of the magnitude difference. T

he formula
b1

b2
= 2.512(m2 − m1) where b1 and b2 represent the brightness and m1 and m2 refer to the magnitudes of the stars being compared. One star has a magnitude of 0.2, and another star has a magnitude of 2.2. What is the ratio of brightness of these two stars?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer: 6.3

Explanation:

The ratio of brightness of two stars is given by:

\frac{b_1}{b_2}=2.512^{(m_{2}-m_{1})}

Where:

b_1 is the brightness of star 1

b_2 is the brightness of star 2

m_{1}=0.2 is the magnitude of star 1

m_{2}=2.2 is the magnitude of star 2

\frac{b_1}{b_2}=2.512^{(2.2-0.2)}

\frac{b_1}{b_2}=2.512^{2}

Finally:

\frac{b_1}{b_2}=6.3 This is the ratio of brightness

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Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a
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Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

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We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

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3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
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Answer:

a) E = 0

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Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

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E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

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E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

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E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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