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kodGreya [7K]
3 years ago
9

A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li

ght projected onto the wall is moving along the wall when the light's angle is 15 degrees from perpendicular to the wall.
Physics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

29.2 ft/s

Explanation:

The distance of the light's projection on the wall

y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,

Angular speed = w = (2π/3)

w = (θ/t)

θ = wt = (2πt/3)

(dθ/dt) = (2π/3)

y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

(dy/dt) = (26π/3) sec² (15°)

(dy/dt) = 29.2 ft/s

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Explanation:

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A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute
Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

  • mass of the bucket, M = 10 kg
  • velocity of pulling the bucket, v = 3m.min^{-1}
  • height of the platform, h = 30 m
  • time taken, t = 10 min
  • rate of loss of water-mass, m = 0.4 kg.min^{-1}

Here, according to the given situation the bucket moves at the rate,

v=3 m.min^{-1}

The mass varies with the time as,

M=(10-0.4t) kg

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x =  3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

W=\int_{0}^{10} [(10-0.4t).g\times 3] dt

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A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

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