A light wave travels through space at a speed of 3 x 108 m/s. If the wavelength of some light wave is 2 x10-3 m, what is the fre
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<span>On what:
f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm
</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
Product of extremes equals product of means:
f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm
</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
Product of extremes equals product of means:
Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
Wₐ = 1500 lb
W_b = 1125 lb
Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
W = mg
m = W / g
mₐ = Wₐ / g
m_b = W_b / g
mₐ = 1500/32 = 46.875 slug
m_b = 125/32 = 35,156 slug
Let's reduce to the english system
v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
v =
we substitute the values
v =
v = 0.559 v₀ₐ - 26.40 (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
M = mₐ + m_b
M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
K₀ = ½ M v²
final point. When they stop
K_f = 0
The work is
W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
Let's write Newton's second law
Axis y
N-W = 0
N = W
the friction force has the expression
fr = μ N
we substitute
-μ W x = Kf - Ko
-μ W x = 0 - ½ (W / g) v²
v² = 2 μ g x
v = Ra (2 0.750 32 17.5
v = 28.98 ft / s
The forces cancel out each other because they are equal and opposite. Therefore the net force is 0 and there will be no acceleration
