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Georgia [21]
3 years ago
8

A person steps off the end of a 3.45 � high diving board and drops to the water below. (a) How long does it take for the person

to reach the water? (b) What is person’s speed on entering the water? (c) What is person’s speed on entering the water if they step off a 12.0 � diving
Physics
1 answer:
777dan777 [17]3 years ago
6 0

Answer:

(a) 0.84 s

(b) 8.22 m/s

(c) 15.33 m/s

Explanation:

u = 0 , h = 3.45 m, g = 9.8 m/s^2

Let time taken to reach the water is t and the velocity at the time of hitting of water surface is v.

(a) Use first equation of motion

s = u t + 1/2 a t^2

3.45 = 0 + 0.5 x 9.8 x t^2

t = 0.84 second

(b) Use first equation of motion

v = u + a t

v = 0 + 9.8 x 0.84 = 8.22 m/s

(c) If h = 12 m

use third equation of motion

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 12

v = 15.33 m / s

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In a certain region of space, the electric field is zero. from this fact, what can you conclude about the electric potential in
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As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce
WINSTONCH [101]

Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

The force of friction has the expression

                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0
3 years ago
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Answer:

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v = 2.913\,\frac{m}{s}

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12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s

1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_{1} \approx 0.128\,m

\Delta s_{2} \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

The maximum height that the block reaches after rebound is:

(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}

y_{max} = 0.829\,m

4 0
3 years ago
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