force of gravitation between two different masses is given by
here in first case we will have
Newton
another case we have
Newton
another case we have
Newton
so the force will be maximum in last case when two 2 kg mass is placed at 1 m distance and minimum force is in second case when 1kg and 2kg mass is placed at 2 m distance
/
c
m
3
.
Part A
Part complete
What is this value in kilograms per cubic meter?
1.35×104
kg/m3
k
g
/
m
3
If the density of mercury is 13.5 g/cm³ in the gram per centimeter cube unit, then the density of the mercury would be 1.35×10⁴ in the kilograms per cubic meter
It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.
By using the above formula for density
ρ = mass / volume
As given in the problem If the density of mercury is 13.5 g/cm³ in the gram per centimeter cube unit then we have to convert its value into kilograms per cubic meter.
1000 grams = 1 kilogram
1 gram = 1/1000 kilograms
=1×10⁻³ kilograms
10⁶ cubic centimeter = 1 cubic meter
1 cubic centimeter = 1/10⁶ cubic meter
= 1× 10⁻⁶ cubic meter
Given the density of the mercury 13.5 g/cm³
13.5 g/cm³ = 13.5 ×10⁻³ kilograms / 1× 10⁻⁶ cubic meter
= 1.35×10⁴ kilograms / cubic meter
Thus, we converted the density of mercury is 13.5 g/cm³ into 1.35×10⁴ kilograms / cubic meter
To learn more about density from here, refer to the link;
#SPJ2
Answer:
DO IY.T
Explanation:
The three conditions that must occur before clouds can form Sufficient water vapor must be present in the air. Water vapor is the invisible gas form of water. It is all around us, even in dry air.
A surface for water vapor to condense on is needed. This is called a cloud condensation nuclei (CCN). CCN can be anything from dust particles to salt crystals.
The air must cool to its dew point. The dew point is the temperature at which the air is saturated with water vapor. When the air reaches its dew point, the water vapor will condense on CCN and form clouds.
In addition to these three conditions, there are other factors that can affect cloud formation, such as the wind and the amount of sunlight.
Read more about cloud formation here:
#SPJ6
Answer:
398.3 m, 334.2 m
Explanation:
The magnitude of the displacement vector is
v = 520 m
And its direction is
measured as north of east.
The x-component of this vector is given by:
While the y-component is given by
The x-component of the displacement vector (east-west direction) is approximately 398 m and the y-component (north-south direction) is approximately 334 m.
Displacement vectors contain both magnitude and direction information. In the case of a displacement vector 520 m long, pointing 40 degrees North of East, we can break it down into x and y components using trigonometric functions, specifically cosine and sine.
For the x-component (east-west direction), we can use the cosine function because it gives us the adjacent side in a right triangle, which is the east-west direction in our case. Therefore, the x-component will be:
520 m * cos(40) = 398m (rounded to nearest unit).
For the y-component (north-south direction), we can use the sine function because it gives us the opposite side in a right triangle, which is the north-south direction in our case. Hence the y-component will be:
520 m * sin(40) = 334 m (rounded to nearest unit).
#SPJ11
What does "observable universe" mean?
Light does not travel instantaneously between points in space. It has a finite speed "c", measured experimentally to be about 3 x 108 meters/second (or about 1.1 x 109 kilometers/hour. Flying at this rate you could get from NYC to Tokyo in about 1/30th of a second.)
Since light takes time to travel, we never actually see the current moment. Looking down at your hand, you do not see it as it is right now, but rather as it was a miniscule fraction of a moment earlier. Now, this interval is so small, given the short distance between your retina and your hand, that the difference is utterly negligible. In fact, bound by Earth's meager scope, the phenomenon isn't really worth mentioning.
The discrepancy becomes significant, however, when exploring much larger distances. Light years, for example. A light year is the distance light travels in one year. If you look at a star that's 50 light years away, you are seeing it as it was 50 years ago. Thus the deeper you peer into space, the farther you are seeing back in time. If this star had exploded 49 years ago, in a spectacular event called a Supernova, we would not know it until 1 more year from now.
Likewise, any event that happened beyond a certain point in the past is unknowable to us if the signal from it hasn't had time to reach us. It is not that our telescopes are too weak, or our instrumentation insensitive. We simply do not yet have access to the information. (No matter how prolific a reader you may be, you'd be hard pressed to read a friend's email if it has yet to arrive in your inbox.)
As a consequence of this limitation, astronomers often refer to the observable universe, a term referring to the volume of space that we are physically able to detect. The question of what lies outside this observable region is a tempting one to ponder. Yet inspiring though it may be, there is a certain futility in such a pursuit.