The path an object takes as it revolves around another object is called an
2 answers:
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Answer:
a) The maximum height of the cannonball is 64.5 m.
b) The cannonball´s speed at maximum height is 50.8 m/s.
Explanation:
The position and velocity vectors of the cannonball can be calculated using the following equations:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity vector at time t.
a) At the maximum height, the vertical component of the velocity vector is 0 (please, see the attached figure and notice that at the maximum height the velocity vector is horizontal).
Knowing this, we can calculate the time at which the cannonball is at its maximum height:
vy = v0 · sin α + g · t
0 m/s = 62.0 m/s · sin 35.0° - 9.81 m/s² · t
- 62.0 m/s · sin 35.0° / -9.81 m/s² = t
t = 3.63 s
Now, we can calculate the y-component of the vector r1 in the figure (r1y):
y = y0 + v0 · t · sin α + 1/2 · g · t²
The cannon is at the same level that the origin of the frame of reference (the ground) so that y0 = 0.
y = 0 m + 62.0 m/s · 3.63 s · sin 35.0° - 1/2 · 9.81 m/s² · (3.63 s)²
y = 64.5 m
The maximum height of the cannonball is 64.5 m
b) To calculate the speed at the maximum height, we can use the equation of the velocity vector:
v = (v0 · cos α, v0 · sin α + g · t)
We already know that the y-component is 0. Then, let´s calculate the x-component of the velocity:
vx = v0 · cos α
vx = 62.0 m/s · cos 35.0°
vx = 50.8 m/s
The vector velocity at maximum height will be:
v = (50.8 m/s, 0)
The speed is the magnitude of the velocity vector:
The cannonball´s speed at maximum height is 50.8 m/s.
(a) the water density is 
, and 1 liter corresponds to a volume of 
. Therefore we can find the mass of the water in the first case:
The amount of heat supplied to the water to raise its temperature by
is
(1)
where
is the specific heat capacity of the water.
Using the data, we find
We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
And so by re-arranging equation (1) we can calculate the new increase of temperature:
(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
And so, the increase in temperature if we use the same amount of heat as before is
The amount of heat supplied to the water to raise its temperature by
where
Using the data, we find
We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
And so by re-arranging equation (1) we can calculate the new increase of temperature:
(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
And so, the increase in temperature if we use the same amount of heat as before is