Answer:
a) 
b) 
c) 
Explanation:
According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.
With this information, we can compute the ball’s starting speed.
a) Let's first assess the horizontal trajectory.
(1)
This gives us our initial equation.
Next, we need to examine the vertical trajectory.
Utilizing 
in our first equation (1)
Now let’s solve for t.
The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.
b) To determine the velocity magnitude just before impact, we must calculate both x and y components.
The computed velocity magnitude is:
c) The ball's angle is:
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Response:
2.5kN.m
Details:
Torque relates directly to the pitch diameter
= Ta/Tb= Da/Db
For 120/Tb= 0.25/0.5
This gives Tb= 2.469kN.m, roughly 2.5kN.m
A hiker proceeds 200 m west and subsequently another 100 m north, resulting in a displacement of 223 m. The direction can be determined using the trigonometric function where sin(angle) = opposite/hypotenuse, yielding an angle of 26.6 degrees. Therefore, the total displacement is 223 m at an angle of 26.6 degrees north of west.
J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
