In an equation f = l^2-d^2/4l the intercept is<br>

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

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

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

5 0

3 years ago

What is an example of velocity?What is an example of velocity?

givi [52]

Speed with direction

5 0

2 years ago

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A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

2 years ago

High pressure center of dry air

Virty [35]

Anticyclone is the high pressure center of dry air

5 0

2 years ago

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Franny drew a diagram to compare images produced by concave and convex lenses.

yanalaym [24]

Answer:

virtual