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sladkih [1.3K]
3 years ago
10

A submarine is built to dive to a depth of 90 m where the pressure in the water is about 1 MPa. The engineers want to build a wi

ndow in the submarine, but the window can only take a force of 10 kN safely. What is the largest surface area window that can be used?
Physics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

<em>The largest surface area window that can be used is</em> A=0.01\ m^2

Explanation:

<u>Pressure</u>

The pressure is defined as force per unit area. The SI unit for pressure is the Pascal (Pa), defined as Newton per square meter.

If a force F acts on a surface area A, the pressure is calculated as:

\displaystyle P=\frac{F}{A}

The pressure at a depth of 90 m is P=1 Mpa= 1,000,000 Pa and the submarine's window can only take a force of F=10 kN=10,000 N.

We need to calculate the largest surface area of the window. We can solve the equation for A:

\displaystyle A=\frac{F}{P}

Substituting:

\displaystyle A=\frac{10,000}{1,000,000}

A=0.01\ m^2

The largest surface area window that can be used is \mathbf{A=0.01\ m^2}

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Answer:

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Explanation:

 From the question we are told that

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     The speed of the transverse wave is v = 8.25 \ m/s

     The amplitude of the transverse wave is A = 5.50 *10^{-2} m

     The wavelength of the transverse wave is \lambda = 0540 m

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          w = vk

Substituting values  

         w = 8.25 * 11.64

        w = 96.03 \ rad/s

The propagation constant k is mathematically represented as

                  k = \frac{2 \pi}{\lambda}

Substituting values

                  k = \frac{2 * 3.142}{0. 540}

                   k =11.64 m^{-1}

Substituting values into the equation for mechanical waves

      y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03  * 0.150))

       y(1.51 , 0.150) = 0.055 m    

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A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
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Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

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            y = y₀ + v_{oy} t - ½ g t²

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            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

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                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

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For u₂

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The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
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