A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a
2 answers:
Answer:
The mass of the second sphere is 15.4 kg
Explanation:
Given;
mass of the first sphere, m₁ = 7.7 kg
initial velocity of the second sphere, u₂ = 0
let mass of the second sphere = m₂
let the initial velocity of the first sphere = u₁
final velocity of the composite system, v = ¹/₃ x u₁ =
From the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
Substitute the given values;
Divide through by u₁
7.7 = ¹/₃(7.7 + m₂)
multiply both sides by 3
23.1 = 7.7 + m₂
m₂ = 23.1 - 7.7
m₂ = 15.4 kg
Therefore, the mass of the second sphere is 15.4 kg
You might be interested in
Hope this helps you.
Answer:
y = 17,89 m
Explanation:
Let us fixate the reference point in top of the building, from where the watermelon is thrown down. We will assume also that the positive axis of our system points up. We describe the watermelon’s motion with the equation:
Clearing the equation so we isolate y we have that:
Making a substitution with the values from the statement we have:
]
So, this skyscraper is about 17,89 m tall; which is not very tall for a skyscraper but who am I to judge. 17,89 m is also the displacement of the watermelon from the point it was thrown down.
I hope everything was clear with my explanation. If I can help with anything else, just let me know. Have an awesome day :D
Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
