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KATRIN_1 [288]
4 years ago
5

9: A post is being driven into the ground. The first strike drives the post 25 inches into the ground. Each additional strike dr

ives the stake 4/5 the distance further into the ground than the previous strike. What is the total distance (to the nearest inch) than the post is driven into the ground after 7 strikes?
Physics
1 answer:
faltersainse [42]4 years ago
3 0

Answer:

99

Explanation:

First strike=25 in

Second strike=4/5*25=20 in

Third strike=4/5*20=16 in

Fourth strike=4/5*16=12.8 in

Fifth strike=4/5*12.8=10.24 in

sixth strike=4/5*10.24=8.192 in

seventh strike=4/5*8.192=6.5536 in

Total=(25+20+16+12.8+10.24+8.192+6.5536) in=98.7856 in

Taking this as a geometric series

Sum(7)=a\frac{(1-r^{7}}{1-r} where a is initial value taken as 25 for this case and r is rate taken as 4/5 or 0.8 in our case

Sum(7)=25\frac {1-0.8^{7}}{1-0.8}=98.7856 in

To the nearest inches, sum is 99 in

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An alternator consists of a coil of area A with N turns that rotates in a uniform field B around a diameter perpendicular to the
matrenka [14]

Answer:

(a): \rm 2\pi f\ NBA\ \sin(2\pi ft)..

(b): 20.94 Volts.

Explanation:

<u>Given:</u>

  • Area of the coil = A.
  • Number of turns of the coil = N.
  • Magnetic field in which the coil is placed = B.
  • The frequency with which the coil is rotating = f.

<h2>(a):</h2>

The magnetic flux linked with a coil is defined as

\phi = N\vec B \cdot \vec A=\rm NBA\cos\theta

where,

  • \vec A is the area vector of the coil, directed along the normal to the plane of the coil.
  • \theta = angle between the magnetic field and the area vector of the coil.

Assuming that the magnetic field is along the normal to the plane of the coil, initially.

Therefore, at any later time t, the angle which the magnetic field makes with the normal to the plane of the coil is is given by

\rm \theta = 2\pi ft.

Therefore, the magnetic flux linked with the coil at any time t is given by

\rm \phi = NBA\cos(2\pi ft).

According to the Faraday's law of electromagnetic induction, the emf induced in the coil is given by

\rm e=-\dfrac{d\phi}{dt}\\=-\dfrac{d(NBA\cos(2\pi ft))}{dt}\\=-NBA\dfrac{d(\cos(2\pi ft))}{dt}\\=-NBA(-2\pi f\sin(2\pi ft))\\=2\pi f\ NBA\ \sin(2\pi ft).

<h2>(b):</h2>

The amplitude of the alternating voltage is the maximum value of the induced emf in the coil, the induced emf in the coil is maximum when \rm \sin(2\pi ft)=1.

Therefore, the amplitude of the alternating voltage is given by

\rm e_o=(2\pi f)NBA\ 1\\=2\pi ft \ NBA.

Given values are:

  • N = 100 turns.
  • A = \rm 10^{-2}\ m^2.
  • B = 0.1 T.
  • f = 2000 rev/min = \rm 2000\times \dfrac1{60}\ rev/s = 33.33\ rev/s.

Putting all these values,

\rm e_o=2\pi \times 33.33\times (100)\times (0.1)\times (10^{-2})=20.94\ Volts.

7 0
3 years ago
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