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sweet-ann [11.9K]
3 years ago
5

Hw2-2 - show all your work, including the equation, express all answers in base units, (m, m/s, m/s2), unless stated otherwise i

n problem. a sprinter accelerates from rest to 10.5 m/s in 1.35 sec. what is her acceleration . . . in m/s2 km/hr2 a certain drag racer can accelerate from 0 to 60.0 km/hr in 5.4 sec. what is its acceleration in m/s2? how far, in meters, will it travel during this acceleration? how much time would it take to go 0.25 km if it could accelerate at this rate the entire time? an automobile can increase its speed uniformly from 25.0 to 55.0 km/h in 30.0 seconds. a bicycle rider uniformly speeds up to 30.0 km/h from rest in 0.5 min. calculate their accelerations. a rocket rises from rest with constant acceleration to an altitude of 85.0 km, at this point it has a speed of 2.8 km/s what is the acceleration of the rocket? how long does the ascent take?
Physics
1 answer:
Papessa [141]3 years ago
4 0
The useful equations here are derived equations for rectilinear motion at constant acceleration:

a = (v - v₀)/t
2ax = v² - v₀²
x = v₀t + 0.5at²

1.  Since it starts from rest, v₀ = 0. Then, v = <span>10.5 m/s and t=1.35 s
     a = (10.5 - 0)/1.35 = 7.78 m/s</span>²
     a = 7.78 m/s²*(1 km/1000 m)*(3,600 s/ 1 h)² = 100,828.8 km/h²

2. v₀ = 0; v = 60 km/h; t=5.4 s 
    a = [60 km/h*(1000 m/1km)*(1 h/3600 s) - 0]/5.4 s = 3.09 m/s²

3. Using the acceleration and velocities in #2, we can determine the distance     by the formula: 2ax = v² - v₀²
    2(3.09 m/s²)x = [60 km/h*(1000 m/1km)*(1 h/3600 s)]² - (0 m/s)²
    Solving for x,
    x = 44.95 m

4. Using acceleration in #2 and v₀ = 0, the time would be
    x = v₀t + 0.5at²
    0.25 km * 1,000 m/1 km = (0)(t) + (0.5)(3.09 m/s²)(t²)
    Solving for t,
    t = 12.72 seconds

5. Acceleration of automobile:
   a = [(55 km/h - 25 km/h)*(1,000 m/1 km)*(1 h/3,600 s)]/30 s
   a = 0.278 m/s²
<span>
   </span>Acceleration of bicycle:
   a = [(30 km/h - 0 km/h)*(1,000 m/1 km)*(1 h/3,600 s)]/(0.5 min * 60 s/1 min)
   a = 0.278 m/s²
<span>
6. x = 85 km or 85,000 m; v</span>₀ = 0 m/s; v = 2,800 m/s
    2ax = v² - v₀²
    2a(85,000 m) = (2,800 m/s)² - 0²
    a = 46.12 m/s²

7. Using the data given and acceleration in #6,
   x = v₀t + 0.5at²
   85,000 m = 0*t + 0.5(46.12 m/s²)(t²)
   Solving for t,
   t = 60.71 seconds
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4 years ago
what is the minimum (nonzero) thickness of the coating for which there is maximum transmission of the light into the rod?
timofeeve [1]

The thickness is 155 nm at t1.

The thickness is 77.3 nm at t2.

The inquiry informs us that the laser light's wavelength is λ=510nm

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The transparent coating's refractive index is nr=1.65

Minimum reflection would be required for maximal light transmission into the rod, and it is mathematically described as

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The highest reflection would occur for minimal light penetration through the rod, and this maximum reflection is mathematically described as

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The complete question is- Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

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brainly.com/question/15487308

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