Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
Answer:
520000 or 520000 pa
Force = 520N
Area of contact = 0.001
Pressure: 520000 or 520000
B) Hope it helps ,Have a nice day :)
Answer: The weight of the air displaced by the balloon is less than the volume of the balloon.
Explanation:
A hot air balloon is a cloth wrap that contains several thousand cubic meters of air inside (a large volume of air). The burner heats the liquid propane to a gaseous state to generate a huge flame, which can reach more than 3 meters, thus heating the air mass inside the balloon. In this way,<u> its density is modified with respect to the air that surrounds it</u>, because the hot air is lighter than the outside air (less dense), causing the balloon to rise and float.
Now, if we know that the density of a body
is directly proportional to its mass
and inversely proportional to its volume
:

We can deduce that <u>by increasing the volume of the body, its density will decrease.</u>
This is proof of <em><u>Archimedes' Principle</u></em>:
<em>A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.</em>
In this case the fluid is the air outside. So, the warm air inside the balloon, being less dense, will weigh less than the outside air and therefore will receive an upward pushing force or thrust that will make the balloon ascend.