Kepler’s three law is the answer. Kepler’s 3 is the amount
of time it takes to orbit the sun is related to size and distance. Kepler’s 3 is one of the planetary motion and
can be stated as all planets move in elliptical orbits, having the sun sits at
one of the foci.
Answer:
<em>The 6000 lines per cm grating, will produces the greater dispersion .</em>
Explanation:
A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.
The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.
For a plane diffraction grating, the angular positions of principle maxima is given by
(a + b) sin ∅n = nλ
where
a+b is the distance between two consecutive slits
n is the order of principal maxima
λ is the wavelength of the light
From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.
In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.
Answer:
density is 
Mg/µL
Explanation:
given data
density of nuclear = 
kg/m³
1 ml = 1 cm³
to find out
density of nuclear matter in Mg/µL
solution
we know here
1 Mg = 1000 kg
so
1 m³ is equal to cm³
and here 1 cm³ is equal to 1 mL
so we can say 1 mL is equal to 10³ µL
so by these we can convert density
density = kg/m³
density = kg/m³ × 
Mg/µL
density = Mg/µL
Answer:
the answer will be option no b plss mark me brainliest
1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.
In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:
The initial velocity of the object before the dropping is 0, so we can reduce the equation to:
We know the height h of the spaceship hovering, and the gravity of Venus is 
. Substituting this values in the equation :
To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:
Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth 
.
