Answer:
T = 1.1285 10⁻² day
Explanation:
For this exercise the forces in the premiere are internal, so the angular momentum is conserved
L₀ = I₀ w₀
L = I w
L₀ = L
I₀ w₀ = I w
Angular velocity and period are related
w₀ = 2π / T₀
w = 2π / T
The moment of inertia of a sphere is
I₀ = 2/5 M R²
I = 2/5 m r²
If we assume that the mass of the star does not change in the transformation
We substitute
2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T
R² /T₀ = r² / T
T = (r / R)² T₀
T = (6.1 / 2.0 104) 37
T = 1.1285 10⁻² day
The imaginary second level is 60 dB more intense than the real level of the caller.
60 dB means a multiplication of 10⁶ = <em>1 million.</em>
That's how many equally-talented callers it would take to be 60 dB louder than him.
The answer is 20 meters would be the water depth started to feel bottom. This is half of the wavelength. The reason behind this is the wave travels half or the partial of the wavelength underneath the surface. For that reason that is where the wave would be starting to feel the ground.
Kepler actually showed that the planets move around the sun in ellipses, not circles. So the answer is false.
Answer:
Given that,
Mass of the Earth m
1
=6×10
24
Kg
Mass of the Moon m
2
=7.4×10
22
kg
Distance between the Earth and the Moon d=3.84×10
5
km=3.84×10
8
m
Gravitational Constant G=6.7×10
−11
Nm
2
/kg
2
Now, by using Newton’s law of gravitation
F=
r
2
Gm
1
m
2
F=
(3.84×10
8
)
2
6.7×10
−11
×6×10
24
×7.4×10
22
F=
14.8225×10
16
297.48×10
35
F=20.069×10
19
F=20.1×10
19
N
Hence, the gravitational force of attraction is 20.1×10
19
N
