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sasho [114]
3 years ago
8

a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?

Physics
1 answer:
Vinil7 [7]3 years ago
5 0

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it.  Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s  ==>  (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out.  If you prefer to use the formula,
the first problem you run into is:  You need to remember the formula !

The formula is        D = 1/2 a T²

                   Distance = (1/2 acceleration) x (time in seconds)²

             Acceleration = (change in speed) / (time for the change)
                                  =        (18 m/s)            /        (12 sec)
                                  =                      1.5 m/s² .

                  Distance  =  (1/2 x 1.5 m/s²) x (12 sec)²
                                  =       (0.75 m/s²)  x  (144 sec²)  =  108 meters .

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den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

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       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

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3 0
3 years ago
What is the kinetic energy of a an 80kg football player running at 8 m/s?
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Then the mass of the football player = 80 kg
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Now we have to put the known data in this equation to find the actual velocity of the footballer.
</span> <span></span>So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
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3 years ago
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Answer:

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So you do

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Ans-7.5

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