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slavikrds [6]
3 years ago
15

A bowling ball is dropped from a height of 24 feet.

Physics
1 answer:
sdas [7]3 years ago
7 0

Answer:

a. H=(1/2)*a*t^2

B.h=16.09ft

C.t= 1.22s

Explanation:

Explanation: Newton's equation of motion. This is called motion under free fall.

S=ut +(1/2)*a*t^2

Let S=H, and a=g

1.

H=ut +(1/2)*g*t^2

H=distance covered in ft(24ft)

u= initial velocity (ft/s)

t= time covered in seconds

g=acceleration due to gravity(ft/s^2)

a=g (32.185ft/s)

When u=0

H=(1/2)*g*t^2

2. H=(1/2)*g*t^2

g=32.185ft/s^2

t=1s

Juxtaposing the values into the equation

H= (1/2)* 32.185*(1)^2

H=16.09ft

3. When the maximum distance covered is 24ft, then using

H=(1/2)*a*t^2

24=0.5*32.165*t^2

24=16.09t^2

1.491=t^2

Find the square root of both sides

t=1.22s

It takes 1.22s to hit the ground

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A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
3 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

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The answer is c you got to look for answers that make sense
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