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e-lub [12.9K]
3 years ago
8

A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis

tance does the tip move in one revolution? What are (b) the tip’s speed and (c) the magnitude of its acceleration? (d) What is the period of the motion?
Physics
1 answer:
olga55 [171]3 years ago
6 0

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = \dfrac{1200}{60}rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

T = \dfrac{1}{f}

T = 1/20 = 0.05 s

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Explanation:

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Show that the energy of a magnetic dipole m in a magnetic field B is U--m B
Juli2301 [7.4K]

Answer:

showm

Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field \vec{B} then the torque exerted by the field on the dipole is

\tau = m\times B

\tau=mBsin\alpha

Now to rotate the dipole in the field to its final position the work required to be done is

U=\int \tau d\alpha

U=\int mBsin\alpha d\alpha

U= -mBcos\alpha

U=-\vec{m\times \vec{B}}

Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

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3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
Ugo [173]

Answer:

f_{e}  = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

    M = M₀ m_{e}

Where M₀ is the magnification of the objective and  m_{e} is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       m_{e} = 25 /  f_{e}

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/f_{e}  

1) Let's look for the focal length of the eyepiece (faith)

     f_{e}  = - L 25 / f₀ M

     M = 400X = -400

     f_{e}  = - 12 25 /0.40 (-400)

     f_{e}  = 1.875 cm

Let's approximate two significant figures

    f_{e}  = 1.9 cm

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3 years ago
Practice questions, will mark brainliest!
andrew-mc [135]

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

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