Answer:
0.000001 kg
Explanation:
because 1 kg equal 1,000,000 milligrams
we take which equals 0.000001 kg
True or False: the longer the hail stone is kicked around by updrafts the larger it gets.
1 answer:
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Answer:
a)
b)
c)
Explanation:
Our most valuable tool in solving this problem will be the conservation of mechanical energy:
That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this mechanical energy will remain constant. (as long as there is no dissipation)
For a point particle, we have that kinetic energy is:
Where m is the mass, and V is the particle's velocity,
Potential energy on the other hand is:
where g is the acceleration due to gravity () and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which
, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.
Let's calculate the mechanical energy just before the ball is dropped:
As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:
That's it, the actual value of m is not important now, as we will see.
Now, what's the potential energy at the bottom? Let's see:
At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set
So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:
And so we have found the velocity of the ball as it hits the floor.
Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:
where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:
Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.
As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:
what's the average speed when the ball is descending?
so the time it takes the ball to go down is:
Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:
and the time it takes to go up is:
When we add both times , we get:
The answer is letter C.
Weight (on Earth) is the force due to the mass of Earth attracting whatever mass is subject of discussion.
The force of attraction between any two masses is called Newton's Law of Universal Gravitation:
is simply a given constant.
If we're at the surface of Eath,
refers to the mass of the Earth, 
to the mass of whatever is on the surface of Earth, and 
to the radius of Earth.
Normally, we define a constant
to be equal to 
; in which 
is the mass of Earth and 
the radius of earth; happens to be around 9.8.
By that, we adapt the Law of Universal Gravitation to objects on the surface of Earth, we call that force Weight.
As you can see, weight is directly proportional to mass, more mass implies more weight.
Weight (on Earth) is the force due to the mass of Earth attracting whatever mass is subject of discussion.
The force of attraction between any two masses is called Newton's Law of Universal Gravitation:
If we're at the surface of Eath,
Normally, we define a constant
By that, we adapt the Law of Universal Gravitation to objects on the surface of Earth, we call that force Weight.
As you can see, weight is directly proportional to mass, more mass implies more weight.