Trace the path of a ray emitted from the tip of the object toward the focal point of the mirror and then the reflected ray that
results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.
1 answer:
Answer: find the attached files for the answer
Explanation:
The reflected ray appears to have originated from the focal point. We should actually draw a vector from the focal point through the point where the incident ray hits the mirror but we shorten the vector so that its starting point is on the mirror, without changing its angle.
Please find the attached files for the solution
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Answer:
greater speed will be obtained for the elastic collision,
Explanation:
To answer this exercise we must find the speed that the sail acquires after each impact.
Let's start by hitting a ball of clay.
The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.
initial instant. before the crash
p₀ = m v₀
where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest
final instant. After the crash
the mass of the candle is M
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v =
for when n balls have collided
v = v₀
Now let's analyze the case of the bouncing ball (elastic)
initial instant
p₀ = m v₀
final moment
p_f = m v_{1f} + M v_{2f}
p₀ = p_f
m v₀ = m v_{1f} + M v_{2f}
m (v₀ - v_{1f}) = M v_{2f}
this case corresponds to an elastic collision whereby the kinetic energy is conserved
K₀ = K_f
½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²
v₁ = v_{1f} v₂ = v_{2f}
m (v₀² - v₁²) = M v₂²
let's use the identity
(a² - b²) = (a + b) (a-b)
we write our equations
m (v₀ - v₁) = M v₂ (1)
m (v₀ - v₁) (v₀ + v₁) = M v₂²
let's divide these equations
v₀ + v₁ = v₂
Let's look for the final speeds
we substitute in equation 1
m (v₀ - v₁) = M (v₀ + v₁)
v₀ (m -M) = (m + M) v₁
v₁ = v₀
we substitute in equation 1 to find v₂
v₂ = v₀ -
v₀
v₂ =
v₂ =
v₂ =
Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.
In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.
Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.
Answer:
180 m
Explanation:
The rock follows a free-fall motion - so the vertical distance covered can be found by using the equation
where
g = 10 m/s^2 is the acceleration due to gravity
t = 6.00 s is the time of the fall
Substituting these data, we find the height of the cliff: