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4 years ago

15

Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa

terfalls as high as 2.87 m. With what minimum speed (in m/s) must a salmon launch itself into the air to clear a 2.87-m waterfall?

1 answer:

patriot [66]4 years ago

5 0

Answer:

The minimum speed is 7.5 m/s.

Explanation:

Given that,

Height = 2.87 m

We need to calculate the minimum speed

Using equation of motion

$v^2=u^2-2gh$

$u=\sqrt{2gh}$

Where, u = minimum velocity

g = acceleration due to gravity

h = height

Put the value into the formula

$u=\sqrt{2\times9.8\times2.87}$

$u=7.5\ m/s$

Hence, The minimum speed is 7.5 m/s.

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4 years ago

Read 2 more answers

A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the accele

ElenaW [278]

(a) $1.23 m/s^2$

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

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

Since u=0 (the block starts from rest), it becomes

$d=\frac{1}{2}at^2$

So by solving the equation for a, we find the acceleration:

$a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2$

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

$W_p = mg sin \theta$

where

m = 3.00 kg is the mass of the block

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

$\theta=33.0^{\circ}$ is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

$F_f = - \mu mg cos \theta$

where

$\mu$ is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

$W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma$

Solving for $\mu$ , we find

$\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50$

(c) 12.3 N

The frictional force acting on the block is given by

$F_f = \mu mg cos \theta$

where

$\mu = 0.50$ is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=33.0^{\circ}$ is the angle of the incline

Substituting, we find

$F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N$

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

$v^2 - u^2 = 2ad$

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

$v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s$

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