Write a sentence or short paragraph that defines and explains an adaptation

A 75.5 kg diver drops from a diving board 10.0 m above the waters surface. Find the divers speed just before he strikes the wate

jenyasd209 [6]

Answer:

Explanation:

<u>Mechanical Energy </u>

The kinetic energy of a body (K) is the capacity of doing work due to its speed. It can be expressed as

$\displaystyle K=\frac{mv^2}{2}$

The potential energy (U) is the capacity of doing work due to its height respect to a certain reference level.

$U=mgh$

The mechanical energy is the sum of both

$\displaystyle E_m=\frac{mv^2}{2}+mgh$

The principle of conservation of mechanical energy states it must remain the same if no external force is acting on it. The diver drops from the diving board, which means its initial speed is zero (and so its initial kinetic energy). Thus, the mechanical energy at the jumping time is

$\displaystyle E_m=mgh=(75)(10)(9.8)=7350\ J$

When the diver is about to get into the water, his height reaches zero and the speed is at maximum. All the potential energy became kinetic energy, so

$\displaystyle \frac{mv^2}{2}=7350\ J$

Rearranging

$\displaystyle v^2=\frac{2(7350)}{75}=196$

$v=\sqrt{196}=14\ m/s$

The final speed of the diver is

$\boxed{v=14\ m/s}$

8 0

3 years ago

When is L larger than XL

polet [3.4K]

-- In Roman Numerals, always ... 50 is larger than 40 .

-- In Algebra, whenever 'X' is less than ' 1 ' .

8 0

2 years ago

Read 2 more answers

What property of an electromagnetic wave changes when the wave is encoded with analog information?

Licemer1 [7]

Answer:

Amplitude and Frequency

Explanation:

Analog signals are composed of continuous waves that can have any values for frequency and amplitude. These waves are smooth and curved.

Radio transmissions are a combination of two kinds of waves: audio frequency waves that represent the sounds being transmitted and radio frequency waves that "carry" the audio information. All waves have a wavelength, an amplitude and a frequency as shown in the figure. These properties of the wave allow it to be modified to carry sound information.

The two most common types of modulation used in radio are amplitude modulation (AM) and frequency modulation (FM). Frequency modulation minimizes noise and provides greater fidelity than amplitude modulation, which is the older method of broadcasting . Both AM and FM are analog transmission systems, that is, they process sounds into continuously varying patterns of electrical signals which resemble sound waves.

7 0

2 years ago

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

Yanka [14]

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) $v_{i}$ = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd equation of motion

$h = v_{y_{f}}t + \frac{1}{2}gt^{2}$

-20 = $v_{y_{f}}$ (4) + 0.5(-9.8)(4)²

$v_{y_{f}}$ (4) = 58.4

$v_{y_{f}}$ = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

$v_{f}$ = $v_{y_{f}}$ / sinθ

$v_{f}$ = 14.6 / sin 60

$v_{f}$ = 16.86 m/s

$v_{x_{f}}$ = $v_{f}$ cosθ

$v_{x_{f}}$ = 16.86 cos 60

$v_{x_{f}}$ = 8.43 m/s

To calculate the horizontal distance

d = $v_{y_{f}}$ t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

$v_{x_{f}}$ = $v_{x_{i}}$

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = ( $v_{y_{f}}$ - $v_{y_{i}}$ ) / t

9.8 = 14.6 - $v_{y_{i}}$ ) / 4

$v_{y_{i}}$ = 24.6 m/s

$v_{i}$ = $\sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }$

$v_{i}$ = $\sqrt{8.43^{2}+24.6^{2}}$

$v_{i}$ = 26 m/s

c)

cos β = $v_{x_{i}}$ / $v_{i}$

β = cos⁻¹ (8.43 / 26)

β = 71.08°

3 0

2 years ago

the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating

VARVARA [1.3K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
$\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}$

4 0

2 years ago