B. kinetic energy increases and potential energy decreases
Answer:
My mom always told me he was just there
Explanation:
Answers: 1) 3 kg m²
2) 2.88 kg m²
Explanation: <u> </u><u>Question 1</u>
I = m(r)²+ M(r)²
I = 1.2 kg × (1 m )² +1.8 kg ×(1 m )²
∴ I = 3 kg m²
<u> </u><u>Question 2 </u>
ACCORDING TO THE DIAGRAM DRAWN FOR QUESTION 2
we have to decide where the center of gravity (G) lies and obviously it should lie somewhere near to the greater mass.<em> (which is 1.8 kg). S</em>ince we don't know the distance from center of gravity(G) to the mass (1.8 kg) we'll take it as 'x' and solve!!
<u>moments around 'G' </u>
F₁ d ₁ = F₂ d ₂
12 (2-X) = 18 (X)
24 -12 X =18 X
∴ X = 0.8 m
∴ ( 2 - x ) = 1.2 m
∴ Moment of inertia (I) going through the center of mass of two masses,
⇒ I = m (r)² +M (r)²
⇒ I = 1.2 × (1.2)² + 1.8 × (0.8)²
⇒ I = 1.2 × 1.44 + 1.8 × 0.64
⇒ I = 1.728 + 1.152
⇒ ∴ I = 2.88 kg m²
∴ THE QUESTION IS SOLVED !!!
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark">
docx
</span>
Answer:
I₂ = 25.4 W
Explanation:
Polarization problems can be solved with the malus law
I = I₀ cos² θ
Let's apply this formula to find the intendant intensity (Gone)
Second and third polarizer, at an angle between them is
θ₂ = 68.0-22.2 = 45.8º
I = I₂ cos² θ₂
I₂ = I / cos₂ θ₂
I₂ = 75.5 / cos² 45.8
I₂ = 155.3 W
We repeat for First and second polarizer
I₂ = I₁ cos² θ₁
I₁ = I₂ / cos² θ₁
I₁ = 155.3 / cos² 22.2
I₁ = 181.2 W
Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized
I₁ = I₀ / 2
I₀ = 2 I₁
I₀ = 2 181.2
I₀ = 362.4 W
Now we remove the second polarizer the intensity that reaches the third polarizer is
I₁ = 181.2 W
The intensity at the exit is
I₂ = I₁ cos² θ₂
I₂ = 181.2 cos² 68.0
I₂ = 25.4 W
Answer:
The parametric equation for the position of the particle is
.
Explanation:
Given that,
The point is

Time t = 3
Velocity 
We need to calculate the parametric equation for the position of the particle
Using parametric equation for position
....(I)
at t = 3,

Put the value into the formula



Put the value of r₀ in equation (I)


Hence, The parametric equation for the position of the particle is
.