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USPshnik [31]
3 years ago
8

Salmon often jump waterfalls to reach their breeding grounds starting downstream, 2.9 meters away from a waterfall .436 meters i

n height, at what minimum speed must a Salmon jumping at an angle of 44.7 degrees leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s^2
Physics
1 answer:
nikklg [1K]3 years ago
4 0

The minimum speed must a Salmon jumping with to leave the water

to continue upstream is 5.79 m/s

Explanation:

At first let us find the two component of the jumping velocity of the fish

1. Horizontal component u_{x} = u cosФ

2. Vertical component u_{y} = u sinФ

where u is the initial velocity and Ф is the angle between the horizontal

and the initial velocity u

→ Ф = 44.7°

→ u_{x} = u cos(44.7)

→ u_{y} = u sin(44.7)

The horizontal distance x is 2.9 meters away from a waterfall

The vertical distance y is 0.436 meters

3. The horizontal distance x = u_{x} t

4. The vertical distance y = u_{y} t + \frac{1}{2} gt²

where g is the acceleration of gravity

→ x = u cos(44.7) t

→ x = 2.9 meters

→ 2.9 = u cos(44.7) t

Divided both sides by u cos(44.7)

→ t = \frac{2.9}{ucos(44.7)} ⇒ (1)

→ y = u sin(44.7) t + \frac{1}{2} gt²

→ y = 0.436 meters , g = -9.81 m/s²

→ 0.436 = u sin(44.7) t - 4.905 t² ⇒ (t)

Substitute (1) in (2) to make the equation of u only

→ 0.436 = u sin(44.7)(\frac{2.9}{ucos(44.7)}) - 4.905 (\frac{2.9}{ucos(44.7)})²

→ 0.436 = 2.9 (\frac{sin(44.7)}{cos(44.7)} - \frac{41.25105}{u^{2}[cos(44.7)]^{2}}

→ 0.436 = 2.8698 - \frac{81.4671}{u^{2} }

Subtract 2.8698 from both sides

→ -2.4338 = - \frac{81.4671}{u^{2} }

Multiply both sides by -1

→ 2.4338 =  \frac{81.4671}{u^{2} }

By using cross multiplication

∴ 2.4338 u² = 81.4671

Divide both sides by 2.4338

→ u² = 33.4732

Take √ for both sides

→ u = 5.79 m/s

<em>The minimum speed must a Salmon jumping with to leave the water</em>

<em>to continue upstream is 5.79 m/s </em>

Learn more:

You can learn more about the equation of trajectory of the projectile in brainly.com/question/2814900

brainly.com/question/5531630

#LearnwithBrainly

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Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

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g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

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The sun appears to move through the background stars. This apparent motion would not exist if:_________.
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Answer:

b) the earth did not orbit the sun

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Question #14
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The decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot bind to another atom, which is why hydrogen has a single atom.

The hydrogen molecule is a form that two hydrogen atoms share their electrons decreasing the total energy of the molecule, this bond has a covalent and hydrogen bonding characteristic.

In a stellar explosion, the energy released increases the energy of the hydrogen atom, for which we have two possibilities:

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When the atoms are thrown into space, the separation between them is so high that it does not allow electrons to be shared and molecules cannot be formed either.

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