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Afina-wow [57]
3 years ago
7

a diver on a board 7.5m above the water walks off the end with a horizontal velocity of 2.3 m/s when they hit the surface of the

water how far away from the edge of the board will they land ?
Physics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

2.846m

Explanation:

The diver is performing projectile motion.

To find x(final), we are going to use the equation x(final) = v(initial)*t + x(initial)

x(initial) = 0

x(final) = ?

v(initial) = 2.3 m/s

we don't know t

To find t we will use y(final) = 1/2*(-9.8)*t^2 + v(initial in the y dir.)*t + y(initial)

- 9.8 in the acceleration in the y dir.

y(final) = 0

y(initial) = 7.5

v(initial in the y dir.) = 0

If we solve for t we get: t = 1.237s

Now we have all the components to solve for x(final) in x(final) = v(initial)*t + x(initial)

x(final) = 2.3*1.237 + 0

x(final) = 2.846m

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So, the wavelength of the light used is 1.28\times 10^{-6}\ m. Hence, this is the required solution.

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
yuradex [85]

Answer: 104.026 m/s=374.49 km/h

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called "drag force" D:  

D={C}_{d}\frac{\rho V^{2} }{2}A (1)

Where:  

C_ {d}=0.7 is the drag coefficient  

\rho=1.21 kg/m^{3} is the density of the fluid (air in this case)

V is the velocity  

A=0.17 m^{2} is the transversal area of the object  

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its weight due to the gravity force W:  

W=m.g (2)

Where:  

m=79.5 kg is the mass of the object

g=9.8 m/s^{2} is the acceleration due gravity  

<h3>So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity: </h3>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g (4)

V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}} (5) This is the terminal velocity

Substituting the known values in (5):

V=\sqrt{\frac{2(79.5 kg)(9.8 m/s^{2})}{(1.21 kg/m^{3})(0.17m^{2}){(0.7)}} (6)

Then:

V=104.026 m/s This is the final velocity in meters per second

Now, let's find the final velocity in kilometers per hour, knowing 1 km=1000 m and 1 h=3600 s:

V=104.026 \frac{m}{s} (\frac{1 km}{1000 m})(\frac{3600 s}{1 h})=374.49 km/h This is the final velocity in kilometers per hour.

4 0
3 years ago
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