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ryzh [129]
3 years ago
8

A train travels 76 kilometers kilometers in 2 hours and then 54 kilometers in 5 hours . What is tuts average speed?

Physics
1 answer:
Elena-2011 [213]3 years ago
7 0
Total distance = 76+54 = 130km
total time = 2+5 = 7hrs
Av. speed = 130/7 = 18.571km/hr = 18.6 km/h ( 3 sig fig)
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Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A
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Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns          Nb = 610 turns

Current = 2.80 A

Average flux through  A  = 300 μWb

Average of flux through B = 90.0  μ Wb

a) L = \dfrac{N \phi}{I}

   L = \dfrac{610\times 90 \times 10^{-6}}{2.80}

   L =19.6 mH

b) inductance of A

   L = \dfrac{N_A \phi_A}{I_A}

   L = \dfrac{430\times 300 \times 10^{-6}}{2.80}

   L =46 mH

c) magnitude of the emf

    \epsilon_B = -L_B\dfrac{dI}{dT}

    \epsilon_B = -(19.6\times 10^{-3})(0.5)

    \epsilon_B = -9.8\times 10^{-3}\ V

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Most asteroids lie between the orbits of
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I believe the answer is B. that's where the asetroid belt is.
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The temperature at one of the Viking sites on Mars was found to vary daily from -90.OF to -5.0 C. Convert these temperatures to
kykrilka [37]

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2 years ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Sedbober [7]

Answer:

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

\omega =\sqrt{\frac{k}{m}}

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

x=A\cos(\omega t)

Finally, the equation of the motion of the system is:

x=(0.088m)\cos(\omega t)

or

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

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3 years ago
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