2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s

a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)


2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
Answer:
166 W
Explanation:
Power is the rate at which work is done.

The work done by Jill is the product of the weight of the pail and the height it moves.
The weight is the product of the mass and acceleration of gravity, <em>g</em>. Taking <em>g</em> as 9.81 m/s², the weight is
<em>W</em> = (6.90 kg)(9.81 m/s²) = 67.689 N
Work done = (67.689 N)(27.0 m) = 1827.603 J
Power = (1827.603 J) ÷ (11.0 s) = 166 W
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final height
is the bomb's initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's final velocity
Knowing this, let's begin with the answers:
<h3>b) Time
</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating
:
(5)
(6)
(7)
<h3>a) Final velocity
</h3>
Since
, equation (3) is written as:
(8)
(9)
(10) The negative sign only indicates the direction is downwards
<h3>c) Range
</h3>
Substituting (7) in (2):
(11)
(12)