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timama [110]
3 years ago
8

A car starts to move from rest and covers a distance of 360m in one minute. Calculate the acceleration of the car.

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0
<h2>The acceleration of car is 0.2 ms⁻²</h2>

Explanation:

When the car moves , the distance covered is calculated by the relation

S = u t + \frac{1}{2} a t²

In this question u = 0 , because car was at rest initially

Thus S =  \frac{1}{2} a t²

here S is displacement and a is the acceleration of car

Therefore  360 =  \frac{1}{2} a ( 60 )²

Because time taken is one minute or 60 seconds

Therefore a = \frac{360x2}{3600}

or a = 0.2 m s⁻²

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Which of the following is not a way that astronomers can find how much dark matter there is in cluster of galaxies?
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Answer:

<em>b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.</em>

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How does deforestation contribute to the greenhouse effect?
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4 years ago
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
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