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Bumek [7]
3 years ago
12

The specific resistance of wire​

Physics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

a material is the resistance offered by a wire of the material which is 1 foot long with a diameter of one MIL. The resistance of a wire is directly proportional to the specific resistance of the material.

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What is this question formula a=V²-U² /2S?​
erik [133]

Answer:

sorry I don't really know about that question.

Explanation:

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2 years ago
The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency?
maria [59]
F=v/wavelength f=1.2/60=50Hz.
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Please help me with this 29 points
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)Give the definition of poverty line as defined by the World Bank.

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2 years ago
A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What
tangare [24]

Answer:

man will move in opposite direction with speed

v_1 = 1.66 \times 10^{-3} m/s

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

P_i = P_f

0 = m_1v_1 + m_2v_2

here we have

0 = (97)v_1 + 0.062(2.6)

v_1 = -\frac{0.1612}{97}

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3 0
2 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
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