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Tresset [83]
3 years ago
10

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. The car continues for 5.00 s at con

stant velocity. How far has the car traveled from its starting point?
Physics
1 answer:
barxatty [35]3 years ago
8 0

Answer:

19.5 m

Explanation:

Using the equation of motion,

For the first 3 seconds,

s = ut +1/2at².......................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance

Note: Since the car starts from rest, u = 0 m/s.

Given: a = 1.0 m/s, t = 3 s, u = 0 m/s

Substitute into equation 1

s = 0×3+1/2(3)²(1)

s = 9/2

s = 4.5

In the remaining five seconds, at a constant velocity,

s' = vt............. Equation 2

Where v = velocity, t = time.

Recall,

v = u+at

v = 0+1(3)

v = 3 m/s.

Also, t = 5 s.

Substitute into equation 2

s' = 3(5)

s' = 15 m.

Total distance = 15+4.5

Total distance = 19.5 m.

Hence the car travels 19.5 m

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Answer:

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Explanation:

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Answer:

1 cm⁻¹ =1.44K  1 ev = 1.16 10⁴ K

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The Planck equation is

          E = h f

Let's start the transformations

     c = f λ = f / ν        

     f = c ν

     E = h f

     E = h c ν

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     h c ν = K T

     T = h c ν  / K =( h c / K) ν

Let's replace the constants

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     c = 3 10⁸ m / s

     K = 1.38  10⁻²³ J / K

 

     v = 1 cm-1 (100 cm / 1 m) = 10² m-1

   

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

     A = h c / K = 1,441 10⁻²

     T =  1.44K

     ν = 103 cm⁻¹ = 103 10² m

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     T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

     T = (h c / K) 1 /λ

     T = 1,441 10⁻²  1 / 1,097 10⁷

     T = 1.3 10⁻⁹ K

    E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

    E = KT

    T = E/K

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Answer:

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