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Bezzdna [24]
3 years ago
13

In 8.4 s a fisherman winds 2.9 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

. The line is being reeled in at a constant speed. Determine the angular speed of the reel.
Physics
1 answer:
alukav5142 [94]3 years ago
3 0

Answer:

The angular speed of the reel is 11.33 rad/s

Explanation:

Given

The fisherman takes t = 8.4 s to wind distance x = 2.9 m into a circle radius of r = 3 cm = 0.03 m

Than the tangencial speed equals the change in the distance to the time

v = \frac{x}{t} = \frac{2.9 m}{8.4 s} = 0.34 \frac{m}{s}

Knowing the tangencial velocity is proportional to the radius r and the angular velocity

v = r*w

w = \frac{0.34 m/s}{0.03 m} = 11.33 \frac{rad}{s}

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Answer:

translation

Explanation:

If a ball has an approximate mass of 0.5 kg, with what force must be kicked to give it an acceleration of 1.5m / s2?

8 0
3 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius r = 0.330

m

Force F = 290 N

Angular acceleration \alpha  = 0.814 \frac{rad}{s^{2} }

From the formula of torque,

 Γ = I\alpha                                        (1)

 Γ = rF                                       (2)

rF = I \alpha

Find momentum of inertia I from above equation,

I = \frac{rF}{\alpha }

I = \frac{0.330 \times 290}{0.814}

I = 117.56 Kg. m^{2}

Find the momentum inertia of disk,

 I = \frac{1}{2}  Mr^{2}

M = \frac{2I}{r^{2} }

M = \frac{2 \times 117.56}{(0.330)^{2} }

M = 2159.045 Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0
4 years ago
PE=?, m=.6kg, g=10m/s2, h=35m<br><br> PLS HELP I NEED THIS DONE
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PE is mgh in this context.
7 0
4 years ago
a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin
KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

where

\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s

Substituting,

\omega' = 0 + (45)(5)=225 rpm

7 0
3 years ago
What does an unconformity in a sequence of rock layers reveal about geologic history?
andreev551 [17]
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7 0
3 years ago
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