Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = 
and filling in:
Δx = 
which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
Relation between electrostatic force and distance is inverse square i.e
1
Fα ----
r^2
Hence if r is tripled, new electrostatic force will be 1/9 times old force.
Answer:
m = 7.48 kg
Explanation:
force (f) = 11,500 N
length of barrel (s) = 1.7 m
angle above the ground = 49.3 degrees
acceleration due to gravity (g) = 9.8 m/s^{2}
initial velocity (u) = 0 m/s
final velocity (v) = 72.3 m/s
mass (m) = ?
force = mass (m) x acceleration (a)
acceleration (a) = force / mass (m)
acceleration (a) = 11500 / m
applying the equation of motion 
, we can get the mass
5227.3 = 0 + 
m = 
m = 7.48 kg
Since the ball is fired horizontally, the initial y velocity is zero and the time to hit the ground is the same as if the ball was simply dropped from the cliff. So you can solve the y position function:
giving a height of 44.1m.
The given final velocity vector tells us that the initial x-directed velocity was about 17m/s.
Answer:
a.
Explanation:
a
The property of air mass include
1) it must be large.
2) it must have relatively uniform properties.
3)it must travel as a recognizable entity.
It must have a warm front at its leading edge, is not necessarily a property of an air mass because Not all air masses have a warm front at their leading edge. There are five types of air masses and different types of the front can be developed.
