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nadya68 [22]
2 years ago
11

What is the economic application of change of state

Physics
1 answer:
Musya8 [376]2 years ago
7 0
It is based on their state like population, literacy and poverty it varies from state to state, country to country
You might be interested in
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
shutvik [7]

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.

The initial height is I = 5 ft and the final height y = 4 ft

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

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

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

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                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

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

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
A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     v_{y} = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

7 0
3 years ago
What is the wavelenght of an earthquake wave if it has a speed of 11 km/s and a frequency of 15 Hz
Zolol [24]

Answer:

The wavelength is about 733 meters.

Explanation:

Use the wavelength-frequency relationship:

\lambda = \frac{v}{f}= \frac{11000 \frac{m}{s}}{15 s^{-1}}\approx733m

The wavelength is about 733 meters.

3 0
3 years ago
How can what you learned be used in the real world? (Ex. At the grovery store, future career, etc). Please be specific and use c
vivado [14]
I could use math to help with baking and measuring how much I should use for a cake or stuff like that, they also show us how to add our money together to help us with stuff.
Math, is helpful with many jobs, stuff learned in school can help you strategize and school also makes you think outside of the box sometimes, which can help with jobs. Also if you want to do a career with fashion designing or engineering, you can use math for that too. School also gives you the ability to read.
5 0
2 years ago
A stuntman of mass 48 kg is to be launched horizontally out of a spring-
Damm [24]

The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s.  The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

  • mv²/2 = ke²/2
  • mv² = ke².................. Equation 1

⇒ Where:

  • m = mass of the stuntman
  • v = velocity of the stuntman
  • k = force constant of the spring
  • e = compression of the spring

⇒ Make v the subject of the equation

  • v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

  • m = 48 kg
  • k = 75 N/m
  • e = 4 m

⇒ Substitute these values into equation 2

  • v = √[(75×4²)/48]
  • v = √25
  • v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

6 0
2 years ago
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