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3 years ago

10

A 78-kg skydiver has a speed of 62 m/s , what is the kinetic energy?

2 answers:

Cloud [144]3 years ago

3 0

Answer:

149,916J

Explanation:

Pneumonic device: Kevin is half-mad and very square

this translates to: KE=(1/2)mv^2 !!

KE=(1/2)(78)(62^2)

KE=(39)(3844)

KE=149,916 Joules

Fofino [41]3 years ago

3 0

Answer:

it's 149916

Explanation:

Formula = 1/2 x m x v^2

so if you put into your calculator 1/2 x 78 x 62^2

you get 149916

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3 years ago

Salmon often jump waterfalls to reach their

erik [133]

Answer:

6.35 m/s

Explanation:

The motion of the salmon is equivalent to that of a projectile, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with constant acceleration ( $g=-9.8 m/s^2$ , acceleration of gravity)

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$v_x = u cos \theta$

where

u = ? is the initial speed

$\theta=32^{\circ}$ is the angle of projection

Then the horizontal distance covered by the salmon after a time t is given by

$d=v_x t =(u cos \theta) t$

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$t=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$h = (u sin \theta) t + \frac{1}{2}gt^2$ (2)

where

$u sin \theta$ is the initial vertical velocity

If we substitute (1) into (2), we get:

$h = (u sin \theta) \frac{d}{cos \theta} + \frac{1}{2}g(\frac{d}{ ucos \theta})^2=d tan \theta + \frac{gd^2}{2u^2 cos^2 \theta}$

We now that in order to reach the breeding grounds, the salmon must travel a distance of

d = 2.02 m

reaching a height of

h = 0.574 m

Substituting these data into the equation and solving for u, we find the initial speed that the salmon must have:

$u =\sqrt{ \frac{gd^2}{2(h-d tan \theta) cos^2 \theta}}=\sqrt{\frac{(-9.8)(2.02)^2}{2(0.574-(2.02)(tan 32))(cos^2(32))}}=6.35 m/s$

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3 years ago

If the speed of sound in air is 343 m/s, what is the shortest wavelength of sound that humans can hear?

GuDViN [60]

F = 1/T = 20,000 so T = 1/20,000
<span>distance = speed * time </span>
<span>L = 343 T </span>
<span>L = 343/20,000 </span>
<span>L =. 01715 meters or about 1.7 centimeters</span>

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