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Natali5045456 [20]
2 years ago
9

Salmon often jump waterfalls to reach their

Physics
1 answer:
PilotLPTM [1.2K]2 years ago
8 0

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

  • height of the waterfall, h = 0.432 m
  • distance of the Salmon from the waterfall, s = 3.17 m
  • angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s

The minimum velocity of the Salmon jumping at the given angle is calculated as;

X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

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Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

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assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

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Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0
3 years ago
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
Andrei [34K]

Answer: A

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2 years ago
A simple machine makes our work easier and faster. why?​
kirill [66]

Answer:

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Explanation:

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4 years ago
Read 2 more answers
PLEASE HELP!!!!
defon

Answer:

v = 2.45 m/s

Explanation:

first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height of cliff = 15 m

Vi = Initial Vertical Velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²

t² = (15 m)/(4.9 m/s²)

t = √3.06 s²

t = 1.75 s

Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,

s = vt

where,

s = horizontal distance covered = 4.3 m

v = original horizontal velocity = ?

Therefore,

4.3 m = v(1.75 s)

v = 4.3 m/1.75 s

<u>v = 2.45 m/s</u>

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