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Nezavi [6.7K]
3 years ago
12

Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2m from a waterfall that is 0.55m tall and jumps

at an angle of 32°. What must be the salmon's minimum speed to reach the waterfall?
Physics
1 answer:
mamaluj [8]3 years ago
7 0
The salmon can travel at a speed of 2894673763462m/s and slide into my microwave! PS- Don't judge the salmon, JUST BECAUSE IT ISN'T ATHLETIC!!!! Worst regards, Vivian Carolina Gartea Chiz Demetreo Perez Sanfrasio Smith
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1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of
lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0
3 years ago
A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th
Usimov [2.4K]

Answer:

\omega_f = 3.584\ rad/s

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed  = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

\theta = \omega_i t + \dfrac{1}{2}\alpha t^2

5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2

\alpha= \dfrac{10}{2.8^2}

       α = 1.28 rad/s²

now, calculation of angular velocity

\omega_f = \omega_i + \alpha t

\omega_f =0 +1.28\times 2.8

\omega_f = 3.584\ rad/s

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0
3 years ago
Which element is found in all the acids
lana66690 [7]
Hydrogen .When acids touches all metal hydrogen gas is emitted .Strong acids is one that can produce a high concentration of hydrogen ions.Hope this helped!
3 0
3 years ago
Read 2 more answers
A 2 kg basketball has a momentum of 4 kg m/s. What is the ball's velocity?
Mumz [18]
That is the answer to the question
I hope this helps you.
Thank you for your question

4 0
2 years ago
A dragster takes off from rest and crosses the finish line 320 m away. If the dragster is able to accelerate at 10LaTeX: \frac{m
Rasek [7]

Answer:

t = 8 s

Explanation:

In order to find the time taken by the dragster we will use equations of motion. Here, we will use second equation of motion:

s = Vi t + (1/2)at²

where,

s = distance covered = 320 m

Vi = Initial Velocity = 0 m/s (Since, dragster starts from rest)

t = time taken = ?

a = acceleration of dragster = 10 m/s²

Therefore,

320 m = (0 m/s)t + (1/2)(10 m/s²)t²

t² = (320 m)(2)/(10 m/s²)

t = √(64 s²)

<u>t = 8 s</u>

3 0
3 years ago
Read 2 more answers
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