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arlik [135]
3 years ago
7

Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength

Physics
1 answer:
9966 [12]3 years ago
3 0
Rarefraction.

Crest- tallest spot on transverse wave.

Trough- shortest point on transverse wave.

Compression - spot on a compressional wave where the waves are closer together.

Rarefraction - spot on a compressional wave where the waves are farther apart.
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Complete this equation that represents the process of nuclear fission
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Answer:

A: 146

B: 56

Explanation:

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A 90-meter train travels at a constant speed of 10 meters per second. How long will it take to cross a 0.06 bridge?
salantis [7]

Answer:

The time taken for the train to cross the bridge is 9.01 s

Explanation:

Given;

length of the train, L₁ = 90 m

length of the bridge, L₂ = 0.06 m

speed of the train, v = 10 m/s

Total distance to be traveled, = L₁ + L₂

                                                  = 90 m + 0.06 m

                                                   = 90.06 m

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Time of motion = 90.06 / 10

Time of motion = 9.006 s ≅ 9.01 s

Therefore, the time taken for the train to cross the bridge is 9.01 s

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A crane takes 2.0 minutes to lift a load to the top of a building. The change in gravitational potential energy of the load is 3
steposvetlana [31]

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A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniform
BartSMP [9]

Answer:

Induced current,I=2.87\times 10^{-3}\ A                                        

Explanation:

Given that,

Area of cross section of the wire, A=200\ cm^2=0.02\ m^2

Time, t = 2.2 s

Initial magnetic field, B_i=0.2\ T

Final magnetic field, B_f=2.8\ T

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=A\dfrac{d(B)}{dt}

\epsilon=A\dfrac{B_f-B_i}{t}

\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}

\epsilon=-0.023\ volts

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

\epsilon= IR

I=\dfrac{\epsilon}{R}

I=\dfrac{0.023}{8}

I=2.87\times 10^{-3}\ A

So, the magnitude of the induced current in the loop of wire is 2.87\times 10^{-3}\ A. Hence, this is the required solution.

7 0
3 years ago
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