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Debora [2.8K]
3 years ago
14

A car moving with uniform acceleration attains speed of 36km/hr in 2 minutes find the acceleration?? ​

Physics
2 answers:
KatRina [158]3 years ago
8 0
<h3 /><h3>Acceleration = 0.083 m/s²</h3><h2>Explanation:</h2>

<h3>Given:</h3>

Initial velocity of car is 0.

Final velocity of car is 36 km/h.

Time taken to attain final speed is 2 minutes.

<h3>To Find:</h3>

What is the acceleration of car ?

<h3>Formula to be used:</h3>

v = u + at

<h3>Solution: </h3>

First change the final velocity from km/h to m/s and also time to seconds.

[ To change km/h to m/s multiply by 5/18 ]

➙ Final velocity = 36(5/18) = 10 m/s.

[ 1 minutes = 60 seconds ]

➙ 2 minutes = 2(60) = 120 seconds.

Now, we have

v = 10 m/s.

u = 0 m/s.

t = 120 s.

  • ⟹ v = u + at
  • ⟹ 10 = 0 + a(120)
  • ⟹ 10 = 120a
  • ⟹ 10/120 = a
  • ⟹ 0.083 m/s² = a

Hence, the acceleration of car is 0.083 m/s².

<h2>__more____info_____</h2>

➟ The rate of change of velocity is called acceleration. It is denoted by a.

➟ The velocity at which motion starts is termed as initial velocity. It is denote by u.

➟ The last velocity of an object after a period of time. It is denoted by v.

Alexandra [31]3 years ago
5 0

\boxed{\large{\bold{\blue{ANSWER~:) }}}}

  • Initial Velocity u = 0

  • Final Velocity v = 36 km/hr = 10 m/s

  • Time t = 2 min = 120 sec

  • Acceleration a = ?

By 1st Equation of motion

<u>we</u><u> </u><u>know that</u><u>, </u>

\boxed{\large{\sf{v \: = u \: + a \: t}}}

<u>according</u><u> </u><u>to the</u><u> </u><u>question</u><u>, </u>

10 = 0 + a \times 120\\\\a \: = \frac{10}{120}\\\\a \: = \frac{1}{12}\\\\a=0.083\: m {s}^{ - 2}

<u>Therefore</u><u>, </u>

Acceleration of the car is 0.083 ms^-2

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     The  radius is r =  3.17 \  mm  =  3.17 *10^{-3} \ m

      The current density is  J =  c\cdot r^2  =  9.00*10^{6}  \ A/m^4 \cdot r^2

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Generally current density is mathematically represented as

          J  =  \frac{I}{A }

Where A is the cross-sectional area represented as

         A  =  \pi r^2

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=>    I  =  J  *  (\pi r^2 )

Now the change in current per unit length is mathematically evaluated as

        dI  =  2 J  *  \pi r  dr

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         I  = 2\pi  \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr

         I  = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr

        I  = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ]  | \left    0.001585} \atop 0}} \right.

        I  = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]

substituting values

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