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dusya [7]
2 years ago
7

Paula ran a total of 72meters directly toward school at a constant velocity. She ran one-third of that distance in 8seconds. Wha

t was her velocity?
Physics
1 answer:
Nezavi [6.7K]2 years ago
3 0

Answer:

3 meters per second

Explanation:

72 divided by 3= 24 meters in 8 seconds

24/8= 3, so 3 meters per second

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   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

3 0
3 years ago
HURRY! PLEASE HELP!!!!<br><br><br> 3. What methods are you using to test this (or each) hypothesis?
Pie

Answer:

How to Test Hypotheses

State the hypotheses. Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. ...

Formulate an analysis plan. The analysis plan describes how to use sample data to accept or reject the null hypothesis. ...

Analyze sample data. ...

Interpret the results.

4 0
3 years ago
A technique in which the muscles are stretched by an outside force is called _____.
geniusboy [140]
A technique in which the muscles are stretched by an outside force is called Passive Stretching 
6 0
3 years ago
The anemometer is spinning and the pressure us high
skelet666 [1.2K]

Answer:

what help you need?????????

3 0
2 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
3 years ago
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