There are already derived equations for rectilinear motion at constant acceleration. For free falling objects, they are further simplified because only the gravitational force is acting on the body. The equations for time and velocity at impact are:
t = √2h/g
v = √2gh
where
h is the height
g is the acceleration due to gravity equal to 9.81 m/s²
1. t = √2(380 m)/9.81 m/s² = 8.8 seconds
2. v = √2(9.81 m/s²)(380 m) = 86.3 m/s
Explanation:
7) Given:
v₀ = 2.0 m/s
v = 0 m/s
t = 3.00 s
Find: Δx
Acceleration isn't included in the problem, so use a kinematic equation that doesn't involve a.
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 2.0 m/s) (3.00 s)
Δx = 3.0 m
8) Given:
v₀ = 0 m/s
v = 5 m/s
t = 4 s
Find: a
Displacement isn't included in the problem, so use a kinematic equation that doesn't involve Δx.
v = at + v₀
5 m/s = a (4 s) + 0 m/s
a = 1.25 m/s²
9) Given:
v_avg = Δx / t
0.5 m/s = 8 m / t
t = 16 s
Start by facing East. Your first displacement is the vector
<em>d</em>₁ = (225 m) <em>i</em>
Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,
<em>d</em>₂ = (350 m) <em>j</em>
Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,
<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )
<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>
The net displacement is
<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃
<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>
and its magnitude is
|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
1 / p = 0.107375
p = 93.13 cm