Answer:
Angle θ = 30.82°
Explanation:
From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ
where;
I_o is the intensity of the polarized wave before passing through the filter.
In this question,
I is 0.708 W/m²
While I_o is 0.960 W/m²
Thus, plugging in these values into the equation, we have;
0.708 W/m² = 0.960 W/m² •cos²θ
Thus, cos²θ = 0.708 W/m²/0.960 W/m²
cos²θ = 0.7375
Cos θ = √0.7375
Cos θ = 0.8588
θ = Cos^(-1)0.8588
θ = 30.82°
Answer:
2200000 = 2.2E6 min for light from Proxima to reach earth
8.3 min from light sun to reach earth
2.2E6/8.3 = 2.56E5 times for light from Proxima
Proxima is about 256,000 times farther away than the sun
Since the sun is about 93,000,000 = 9.3E7 miles from earth
Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away
Note - the speed of light is
3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given
Answer:
F=248.5W N
Explanation:
Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

Since we start counting at 0s and the ball departs from rest, this is just 
So we can write:

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

Substituting our values:

Where the average force exerted has been written it terms of the tennis ball's weight W.
I think the answer is c chemical change
Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec