(The answer is probably obvious for some but I'm not big on science. Not my cup of English-..I'm implying Enlish is more my styl
e...but yeah..help-)
This ray shows the image formed when a candle is placed in front of a curved mirror. Which of the following best describes the light ray shown in red?
A. The ray travels from the focal point to the image and continues in a straight line.
B. The ray travels toward the vertex and is reflected at an equal angle on the opposite side of the axis.
C. The ray travels parallel to the axis and is reflected as if it came from the focal point.
D. Tha ray travels toward the focal point and is reflected parallel to the axis.
This ray shows the image formed when a candle is placed in front of a curved mirror. Which of the following best describes the light ray shown in red?
A. The ray travels from the focal point to the image and continues in a straight line.
B. The ray travels toward the vertex and is reflected at an equal angle on the opposite side of the axis.
C. The ray travels parallel to the axis and is reflected as if it came from the focal point.
D. Tha ray travels toward the focal point and is reflected parallel to the axis.
1 answer:
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Answer:
So energy is not conserved and inelastic shock
Explanation:
In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.
Let's find the kinetic energy just before the crash
K₀ = ½ m vₓ₀²
After the crash we can use the law of conservation of energy
Starting point. Right after the crash, before starting to climb
Em₀ = K = ½ (m + M) v₂²
Final point. At the maximum height of the pendulum
= U = (m + M) g h
Where m is the mass of the bullet and M is the mass of the pendulum
Em₀ = Em_{f}
½ (m + M) v₂² = (m + M) g h
v₂ = √ 2g h
Now we can calculate the final kinetic energy
K_{f} = ½ (m + M) v₂²²
K_{f} = ½ (m + M) (2gh)
The relationship between these two kinetic energies is
K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)
K₀ / K_{f} = m / (m + M) vₓ₀² / 2 g h
We can see that in this relationship the Ko> Kf
So energy is not conserved and inelastic shock