When the forces acting on a body are balanced, their effect
\on the body's motion is the same as if no forces at all are
acting on it, and its velocity can't change. It continues moving
in a straight line at constant speed (which may be zero).
Answer:
159241.048 cm³/s
Explanation:
r = Radius = 3×height = 3h
h = height = 16 cm
Height of the pile increases at a rate = 
Differentiating with respect to time
∴ Rate is the sand leaving the bin at that instant is 159241.048 cm³/s
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is
<em>x</em> = (7.2 m/s) <em>t</em>
The object's height <em>y</em> at time <em>t</em> is
<em>y</em> = 9.4 m - 1/2 <em>gt</em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is
<em>v</em> = -<em>gt</em>
(a) The object hits the ground when <em>y</em> = 0:
0 = 9.4 m - 1/2 <em>gt</em>²
<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)
<em>t</em> ≈ 1.92 s
at which time the object's vertical velocity is
<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s
(b) See part (a); it takes the object about 1.9 s to reach the ground.
(c) The object travels a horizontal distance of
<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m
Vector quantities have both magnitude and direction. Distance is a scalar quantity. It refers only to how far an object has traveled. For example, 4 feet is a distance; it gives no information about direction.
