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ivolga24 [154]
2 years ago
15

. (a) A world record was set for the men's 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coast

ed" across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
Physics
1 answer:
cricket20 [7]2 years ago
7 0

Answer:

a) The acceleration of Usain Bolt was 2.98 m/s².

His maximum speed was (v = 2.98 m/s² · 3.00 s) 8.94 m/s.

b) The acceleration was 3.21 m/s² and the maximum speed was 9.63 m/s

Explanation:

The equation for the position of Bolt during the first 3 s is as follows:

x = x0 + 1/2 · a · t²

Where:

x =  position at time t

a = acceleration

t = time

x0 = initial position

After that time, the position is:

x = x0 + v · t

Where "v" is the velocity

During the acceleration, the velocity is:

v = a · t

The maximum speed reached by Bolt is then:

v = a · 3.00 s

The distance traveled during the first 3 seconds is:

x = x0 + 1/2 · a · t²   (x0 = 0)

x = 1/2 · a · (3.00 s)²

x = 4.50 s² · a

Then, in the equation of the position after the first 3 s, we can replace the initial position for the position traveled during the first 3 s and the velocity for the velocity reached in those 3 s.

x = x0 + v · t

x0 = 4.50 s² · a

v = a · 3.00 s

Then:

x = 4.50 s² · a + 3.00 s · a ·t  

At t = 9.69 s, x = 100 m. Then:

100 m =  4.50 s² · a + 3.00 s · a · 9.69 s

100 m = 33.6 s² · a

100 m / 33.6 s² = a

a = 2.98 m/s²

The acceleration of Usain Bolt was 2.98 m/s².

His maximum speed was (v = 2.98 m/s² · 3.00 s) 8.94 m/s.

b) The calculations are the same only that in the final step we have to use x = 200 m instead of 100 m and t = 19.30 s instead of 9.69 s:

200 m =  4.50 s² · a + 3.00 s · a · 19.30 s

200 m = 62.4 s² · a

200 m / 62.4 s² = a

a = 3.21 m/s²

v = 3.21 m/s² · 3.00 s = 9.63 m/s

The acceleration was 3.21 m/s² and the maximum speed was 9.63 m/s

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The magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2. 1 g that has been charged to -3. 0 NC is 6.86 × 10^{6} N/C.

An electric field is the physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field of a system of charged particles.

It is given that,

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E = 6860000 N/C

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A brick is released with no initial speed from the roof of a building and strikes the ground in 1.80 s , encountering no appreci
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man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
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Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

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Answer:

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