The effective temperature of a star is relative to the
fourth root of the luminosity and is contrariwise proportional to the square
root of the radius.
L = k R² T⁴
If the radius remains continuous, while the luminosity doubles, the temperature
must increase by a factor of the fourth root of two.
If L → 2L, then T → 1.189207115 T
So the answer is approximately 1.19 times.
K.E = 1/2 mv²
100 = 1/2 (2 )(v)²
100 = both 2 cancel (v)²
Taking Square root on b/s
√100 = √v²
10 = v
I would have to say the answer it false
Answer:
2.72 m
Explanation:
wavelength of sound λ = velocity / frequency
= 340 / 1200
= .2833 m
Distance of point of first constructive interference
= λ D / d ( D is distance of the screen and d is distance between source of sound.
Here D = 12.5 m
d = 1.3 m
λ D / d= ( .2833 x 12.5) / 1.3
= 2.72 m
Distance of point of first constructive interference = 2.72 m
Answer:
The answer is B. less permeability.
Explanation:
