Answer:
Because they are too small
Explanation:
The radius of the Earth is approximately 6370 km. This means that distances between farther places in the Earth, e.g. between two cities, is usually measured using the kilometer. Shorter distances are even measured using the metre.
On the contrary, distances in space are much, much greater. For example, the average distance between the Earth and the Sun is approximately 150 million kilometers, which means . For farther objects, it is even unpractical to measure the distance in million km. For example, the closest star to the Sun, Proxima Centauri, is 4.2 light-years from us (the light-year is the distance covered by the light in one year: 1 light year is equivalent to ).
More distant objects, such as galaxies, are even measured using another unit, the parsec: 1 parsec (1 pc) corresponds to .
Answer:
The ratio of T2 to T1 is 1.0
Explanation:
The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.
Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.
Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively
P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively
M = mass of the Sun
m = mass of the spheres, equal masses.
For the first sphere that is distance R from the sun.
F₁ = (GmM/R²)
P₁ = (k/R²)
T₁ = (F₁/P₁) = (GmM/k)
For the second sphere that is at a distance 2R from the sun
F₂ = [GmM/(2R)²] = (GmM/4R²)
P₂ = [k/(2R)²] = (k/4R²)
T₂ = (F₂/P₂) = (GmM/k)
(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0
Hope this Helps!!!
From that graph of the force of the sun on a comet, you can see that the force increases as the distance decreases. (A)
Answer:
true
Explanation:
i've read about it in a bunch of different articles and stuff
Answer:
Explanation:
In an electric field E force on charge q
F = Eq , acceleration a = Eq / m
a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷
= 636.16 x 10⁸ m /s²
b )
initial velocity u = 0
final velocity v = 1.46 x 10⁶ m/s
v = u + at
1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t
t = 2.29 x 10⁻⁵ s
c )
s = ut + 1/2 a t²
= 0 + .5 x 636.16 x 10⁸ x ( 2.29 x 10⁻⁵ )²
= 1668 x 10⁻²
= 16.68 m
d )
Kinetic energy = 1/2 m v²
= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²
= 1.78 x 10⁻¹⁵ J .