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Inessa [10]
3 years ago
15

A fish is hooked on a line that is wound around the spool of a fishing reel. The reel has a mass of 652 g and a diameter of 10 c

m. After 5 seconds, the reel is spinning with an angular velocity of 306 rad/s Assuming that the fish was pulling on the line with a constant force, what was the magnitude of this force?
Physics
1 answer:
iren2701 [21]3 years ago
3 0

The magnitude of the force exerted by the fish is 3,052.5 N.

The given parameters;

  • mass of the reel, m = 652 g = 0.652 kg
  • diameter of the reel, d = 10 cm
  • radius of the reel, r= 5 cm = 0.05 m
  • time of motion, t = 5 s
  • angular speed of the reel, ω = 306 rad/s

According to Newton's third law, the force exerted by the reel is equal in magnitude to the force fish exerted on the reel.

The centripetal force of the reel = force exerted by the fish

F_c = m\omega^2 r\\\\F_c = 0.652 \times (306)^2 \times 0.05\\\\F_c = 3,052.5 \ N

Thus, the magnitude of the force exerted by the fish is 3,052.5 N.

Learn more here: brainly.com/question/14905888

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A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
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Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

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v_{avg}=\dfrac{y_{tot}}{t}

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\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

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