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
Answer:
the velocity of the boats after the collision is 4.36 m/s.
Explanation:
Given;
mass of fish, m₁ = 800 kg
mass of boat, m₂ = 1400 kg
initial velocity of the fish, u₁ = 12 m/s
initial velocity of the boat, u₂ = 0
let the final velocity of the fish-boat after collision = v
Apply the principle of conservation of linear momentum for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
800 x 12 + 1400 x 0 = v(800 + 1400)
9600 = 2200v
v = 9600/2200
v = 4.36 m/s
Therefore, the velocity of the boats after the collision is 4.36 m/s.
final velocity = initial
velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration
distance = (initial velocity x time) +
1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
<span>distance = 0.18 m</span></span>
Answer:
1885.2 ohms
Explanation:
Step one:
given data
L=5H
f=60Hz
Required
The inductive reactance of the inductor
Step two:
Applying the expression
XL= 2πfL
substitute
XL=2*3.142*60*5
XL=1885.2 ohms
