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igomit [66]
2 years ago
15

What do you mean by focus in science

Physics
1 answer:
Tanzania [10]2 years ago
4 0
“a point at which rays of light, heat, or other radiation meet after being refracted or reflected.” Meaning multiple light rays or heat (and other forms of radiation) are all being refracted or reflecting to a certain point
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Identify the equation used to calculate the perpendicular force (F⊥) acting on a block on an inclined plane.
pychu [463]
Using geometrical arguments, we can see that the angle of the inclined plane \theta is equal to the angle between Fg and the perpendicular force.

But the perpendicular force is the projection of Fg along the perpendicular axis, and Fg=mg, so the correct answer is
<span>C) F=mg cosΘ </span>
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3 years ago
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HELP ASAP PLEASE - Which best describes how the arrangement of the sun, moon, and the earth affect the range of the tides during
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The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.5 m. The block has a thermal conductivity of 200 J/(s·m·C˚). In d
Dennis_Churaev [7]

Answer:

Q_p=18000\ J

Q_o=8000\ J

Q_g=72000\ J

Explanation:

Given:

  • Length of block, l=3L_o=3\times 0.5=1.5\ m
  • Breadth of block, b=L_o=0.5\ m
  • height of block, h=2L_o=2\times 0.5=1\ m
  • Thermal conductivity of the block, k=200\ W.m.^{\circ}C
  • Temperature on the hotter side, T_H=37^{\circ}C
  • temperature on the cooler side, T_L=7^{\circ}C
  • time for which the heat flows, t=4\ s

<u>REFER THE ATTACHED IMAGE FOR THE REFERENCE</u>

<em>The rate of heat flow using </em><em>Fourier's law</em><em> of conduction is given as:</em>

\frac{Q}{t}=k.A.\frac{dT}{dx}

<u>Now the amount heat flow perpendicular to the pink surface:</u>

\frac{Q_p}{4}=200\times (0.5\times 1.5).\frac{30}{1}

Q_p=18000\ J

<u>Now the amount heat flow perpendicular to the orange surface:</u>

\frac{Q_o}{4}=200\times (0.5\times 1).\frac{30}{1.5}

Q_o=8000\ J

<u>Now the amount heat flow perpendicular to the green surface:</u>

\frac{Q_g}{4}=200\times (1.5\times 1).\frac{30}{0.5}

Q_g=72000\ J

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An object 4cm high is placed 15 cm from a convex lens of focal length 18cm. Draw a ray diagram and find the position, size, and
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