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
Longitudinal wave hope this is right :))))))))))
<h2>
Answer:</h2>
-310J
<h2>
Explanation:</h2>
The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e
ΔE = Q + W ----------------------(i)
If heat is released by the system, Q is negative. Else it is positive.
If work is done on the system, W is positive. Else it is negative.
<em>In this case, the system is the balloon and;</em>
Q = -0.659kJ = -695J [Q is negative because heat is removed from the system(balloon)]
W = +385J [W is positive because work is done on the system (balloon)]
<em>Substitute these values into equation (i) as follows;</em>
ΔE = -695 + 385
ΔE = -310J
Therefore, the change in internal energy is -310J
<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>
<em />
<em />
Answer:
<u>411.84 kg m/s</u>
Explanation:
Formula :
<u>Momentum = mass × velocity</u>
<u />
=========================================================
Given :
⇒ mass = 26.4 kg
⇒ velocity = 15.6 m/s
=========================================================
Solving :
⇒ Momentum = 26.4 × 15.6
⇒ Momentum = <u>411.84 kg m/s</u>
Answer:
Speed of the aircraft = 36.64 m/s
Explanation:
Consider the vertical motion of the projectile,
We have equation of motion s = ut+0.5at²
Let the velocity of plane be v.
Vertical velocity = vsin55 = Initial velocity of projectile in vertical direction = u
acceleration, a = 9.81 m/s²
displacement , s = 554 m
time, t = 8 s
Substituting,
554 = vsin55 x 8 +0.5 x 9.81 x 8²
v = 36.64 m/s
Speed of the aircraft = 36.64 m/s
