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.
<span> Using conservation of energy
Potential Energy (Before) = Kinetic Energy (After)
mgh = 0.5mv^2
divide both sides by m
gh = 0.5v^2
h = (0.5V^2)/g
h = (0.5*2.2^2)/9.81
h = 0.25m
</span>
Answer:
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Explanation:
For an atom to achieve a stable electron configuration, it needs 8 valence electrons. Since this atom has 6 valence electrons, it needs 2 more to achieve a stable electron configuration.
8-6=2