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zzz [600]
4 years ago
8

An alligator swims to the left with a constant velocity of 5 S/M

Physics
1 answer:
Goshia [24]4 years ago
8 0

Answer:

t = 0.85[s]

Explanation:

To solve this problem we must make a complete description of this. By doing an internet search, we find the description of this problem as if of the question.

<u>Description</u>

<u />

"An alligator swims to the left with a constant velocity of 5 m s when the alligator season a bird straight ahead the alligator speeds up with a constant acceleration of 3 m/s^2 leftward until it reaches a final velocity of 35 Ms left work how many seconds does it take the alligator to speed up from 5 m/s to 35 m/s".

To solve this problem we must identify the initial data:

v0 = initial velocity = 5 [m/s]

a = acceleration = 3 [m/s^2]

vf = final velocity = 35[m/s]

t = time = ?

Using the following kinematic equation, we can find the time that is required.

v_{f}=v_{0}+a*t\\35=5+35*t\\t=\frac{35-5}{35} \\t=0.85[s]

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19.6133 newton.

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Which of the following is a true statement about energy?
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A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant
djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference,\frac{dC}{dt}=12 cm/s

Magnetic field,B=0.5 T

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We know that induced amf,E=\frac{Bd(A)}{dt}

Area of circular coil,A=\pi r^2

E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}

Circumference of circular coil,C=2\pi r

165=2\pi r

r=\frac{165}{2\pi}

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s

Radius of coil at time t=9 s

r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m

1 m=100 cm

E=-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V

Magnitude of induced emf=0.005 V

4 0
3 years ago
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A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a
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erastova [34]

Answer:

The intensity is  I_2  =  1654 \ W/m^2

Explanation:

From the question we are told that

    The intensity of the unpolarized light is  I_o  =  4000 \ W/m^2

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substituting values

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Then the intensity of  incident light emerging from the second polarizer is mathematically represented by Malus law as

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substituting values

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