What forms when air masses with different temperatures meet?
2 answers:
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The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.
Answer:
The arrow will hit 112.07 m from the point of release.
Explanation:
The equation for the position of an object in a parabolic movement is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position
v0 = initial velocity
α = launching angle
y0 = initial vertical position
t = time
g = acceleration due to gravity
We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).
Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t² (g is downward)
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s ( the negative value is discarded)
Now, with this time we can calculate the horizontal distance:
x = x0 + v0 · t · cos α (x0 = 0, the same as y0)
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
