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3 years ago

Salmon often jump waterfalls to reach their

Physics

1 answer:

____ [38]3 years ago

5 0

Answer:

Using the range formula R = v^2 sin 2 theta / g

or v^2 = R * g / sin 86.4

v^2 = 3.14 m * 9.81 m/s2 / .998

v^2 = 30.9 m^2 / s^2

v = 5.56 m/s

This hasn't really proved the question - this would give

vy = 5.56 * sin 43.2 = 3.81 m/s

vx = 5.56 * cos 43.2. = 4.05 m/s

t = 1.57 / 4.05 = .387 sec to reach the waterfall

h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls

There seems another way to do this

vy / vx = tan 43.2 vy = .939 vx

h = vy t - 1/2 g t^2 and t = 1.57 / vx

h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2

Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s

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For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

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A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

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v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

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Therefore, the velocity after

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$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

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For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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