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Anna007 [38]
2 years ago
12

You're caught running a red light on Hwy 316. Attempting to impress the skeptical patrol officer with your physics knowledge, yo

u claim that you were traveling so fast that the red light (685 nm) appeared yellow (590 nm) to you. How fast would you have been traveling (in mi/hr) if that had been the case?
Physics
1 answer:
timurjin [86]2 years ago
3 0

Answer:

Explanation:

We shall apply here Doppler's effect in optics . The formula is as follows

\frac{\triangle\lambda }{\lambda } = \frac{v}{c}

Δλ is change in wavelength , λ is original wavelength , v is velocity and c is velocity of light

Δλ = 685 - 590 = 95 nm

λ = 685

95 / 685 = v / 3 x 10⁸

v = .416 x 10⁸ m / s

= 4.16 x 10⁷ m /s

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What location on earth experiences the most change the number of daylight hours?
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3 years ago
A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part
borishaifa [10]

Answer:

v₂ = 7/ (0.5)= 14 m/s

Explanation:

Flow rate of the fluid

Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.

The formula for calculated the flow rate is:

Q= v*A Formula (1)

Where :

Q is the Flow rate (m³/s)

A is the cross sectional area of a section of the pipe (m²)

v is the speed of the fluid in that section (m/s)

Equation of continuity

The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:

Q₁= Q₂

Data

A₁ = 2m² : cross sectional area 1

v₁ = 3.5 m/s : fluid speed through A₁

A₂ = 0.5 m² : cross sectional area 2

Calculation of the fluid speed through A₂

We aply the equation of continuity:

Q₁= Q₂

We aply the equation of Formula (1):

v₁*A₁= v₂*A₂

We replace data

(3.5)*(2)= v₂*(0.5)

7 = v₂*(0.5)

v₂ = 7/ (0.5)

v₂ =  14 m/s

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A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
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  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

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  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
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