Answer: -33.3 * 10^9 C/m^2( nC/m^2)
Explanation: In order to solve this problem we have to use the gaussian law, the we have:
Eoutside =0 so teh Q inside==
the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.
Then we simplify and get
σ= -4.6/(2*π*b)= -33.3 nC/m^2
For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as
L=2.7*10^30w
<h3> What is its luminosity?</h3>
Generally, the equation for the luminosity is mathematically given as
L=4*\pi^2*b
Therefore
L=4*\pi^2*b
L=4* \pi *(2.83*10^{18})*2.7*10^{-8}
L=2.7*10^30
In conclusion, the luminosity
L=2.7*10^30w
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brainly.com/question/25770676
If they have self motivation or others motivation, they will show their full potential.
Answer:
B. and D. would be my best guess.
Explanation:
The reason why is because if you lower the resistance, the voltage will be higher, and if you higher the voltage, the resistance would be lower and the voltage would higher.
Answer:
a) p = 4.96 10⁻¹⁹ kg m / s
, b) p = 35 .18 10⁻¹⁹ kg m / s
,
c) p_correst / p_approximate = 7.09
Explanation:
a) The moment is defined in classical mechanics as
p = m v
Let's calculate its value
p = 1.67 10⁻²⁷ 0.99 3. 10⁸
p = 4.96 10⁻¹⁹ kg m / s
b) in special relativity the moment is defined as
p = m v / √(1 –v² / c²)
Let's calculate
p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)
p = 4.96 10⁻¹⁹ / 0.141
p = 35 .18 10⁻¹⁹ kg m / s
c) the relationship between the two values is
p_correst / p_approximate = 35.18 / 4.96
p_correst / p_approximate = 7.09
