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Ainat [17]
3 years ago
13

A pendulum is released from rest at point A. If the horizontal line represents the reference point, the pendulum has energy at p

oint B.
Physics
2 answers:
PSYCHO15rus [73]3 years ago
7 0

b i think

hope it helps ^^

Rzqust [24]3 years ago
6 0
Only kinetic

______________

Okay?


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The rate which light flows through a given area of space is referred to as ita
Brilliant_brown [7]
Hey there! 

The correct answer to your question is: Intensity

The rate which light flows through a given area of space is referred to as its intensity. I<span>ntensity and wavelength are two factors which contribute to light energy.</span>
Thank you!
7 0
2 years ago
Could I get help plz
Sloan [31]

Answer:

440hz

Explanation:

saxophone a plays at 430hz and a frequency of 5 beats per second can be heard so saxophone b is playing at a frequency 10hz louder than saxophone A making it 440hz

3 0
2 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
Alex, parked by the side of an east-west road, is watching car P, which is moving in a westerly direction. Barbara, driving east
Alexxandr [17]

Answer:

v_{PB} = 130\ km/h

Explanation:

Since, Alex is at rest. Therefore, the speed measured by him will be the absolute speed of car P. Therefore, taking easterly direction as positive:

Absolute\ Velocity\ of\ Car\ P = v_{P} = -78\ km/h

And the absolute velocity of Barbara's Car is given as:Absolute\ Velocity\ of\ Barbara's\ Car = v_{B} = 52\ km/h

Now, for the velocity of Car p with respect to the velocity of Barbara's Car can be given s follows:

Velocity\ of\ Car\ P\ measured\ by\ Barbara = v_{PB} = v_{B}-v_{P}\\\\v_{PB} = 52\ km/h-(-78\ km/h)

v_{PB} = 130\ km/h

6 0
2 years ago
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