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BartSMP [9]
3 years ago
15

This is for science ill give you brainliest

Physics
2 answers:
kiruha [24]3 years ago
8 0

The<em> last choice</em> is the only one on the list that has the objects in the right order of sizes . . . the Milky Way galaxy <em>is</em> bigger than the solar system.

But you have to think about this a little more, and try to understand something.  (This is going to be confusing, but that's just the way it is.  There's no way for a human mind to really wrap itself around the REAL distances in space.)

If the solar system shrank down to the size of a golf ball, then the Milky Way would be shrunken down to about 6.5 times the size of the real solar system ! !

The football doesn't quite get that idea across.

Allisa [31]3 years ago
4 0

The golf ball represents the solar system and the football represents the Milky Way. Our solar system is <em>a lot</em> smaller than the Milky Way, just like how a golfball is smaller than a football.

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I would think the answer is c.

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Explanation:

1.

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Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

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4.85 m/s / 0.3 s = 16.2 m/s^2

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
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Answer:

a

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b

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   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

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When an action potential reaches the axon terminal, it depolarizes the membrane and opens voltage-gated Na+ channels. Na+ ions enter the cell, further depolarizing the presynaptic membrane.

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