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antiseptic1488 [7]
3 years ago
14

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

trobe light. what radiation pressure is produced on the cardboard by the light?
Physics
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

The value of radiation pressure is 2.933 \times 10^{-8} Pa

Explanation:

Given:

Intensity I = 8.8 \frac{W}{m^{2} }

Area of piece A = 2.1 \times 10^{-4} m^{2}

From the formula of radiation pressure in terms of intensity,

   P = \frac{I}{c}

Where P = radiation pressure, c = speed of light

We know value of speed of light,

 c = 3 \times 10^{8} \frac{m}{s}

Put all values in above equation,

  P = \frac{8.8}{3 \times 10^{8} }

  P = 2.933 \times 10^{-8} Pa

Therefore, the value of radiation pressure is 2.933 \times 10^{-8} Pa

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Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, a
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Answer:

   F = 196 N

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For this exercise we will use Newton's second law,  we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

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In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

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3 years ago
A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac
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Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

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y-6=-\frac{6}{7}x

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\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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