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victus00 [196]
4 years ago
9

Suppose you wanted to start a fire using sunlight and a mirror. Which of the following statements is most accurate? A) It would

be best to use a plane mirror. B) It would be best to use a concave mirror, with the object to be ignited positioned halfway between the mirror and its center of curvature. C) It would be best to use a concave mirror, with the object to be ignited positioned at the center of curvature of the mirror. D) It would be best to use a convex mirror. E) One cannot start a fire using a mirror, since mirrors form only virtual images.
Physics
1 answer:
Maru [420]4 years ago
6 0
<h2>Answer: B) It would be best to use a concave mirror, with the object to be ignited positioned halfway between the mirror and its center of curvature. </h2>

Explanation:

To start a fire it is more optimal to use a concave mirror than a plane mirror. This is because the concave mirror allows concentrating sunlight at a point (the focal point) on an object that acts as fuel and ignite the fire there.

For this it is necessary the object to be  positioned between the center of curvature of the mirror and the mirror (its focus). Thus the rays of the Sun, when converging on the focus, will heat the object and make it burn.

Hence, the correct option is B.

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A diffraction grating with 140 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i
Degger [83]

Answer:

0.003181 radians

0.003005 radians

Explanation:

Number of slits = 140 /cm

λ = Wavelength = 434 nm = 434×10⁻⁹ m

m = 3 Third order spectrum

Space between slits

d=\frac{0.01}{140} =7.14\times 10^{-5}\ m

Now,

dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 434\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.018228\\\Rightarrow \theta=0.01823^{\circ}=0.01823\times \frac{\pi}{180}=0.003181 radians

0.003181 radians

When λ = 410 nm = 410×10⁻⁹ m

dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 410\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.01722\\\Rightarrow \theta=0.01722^{\circ}=0.01722\times \frac{\pi}{180}=0.003005 radians

0.003005 radians

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there are two types of waves: electromagnetic and mechanical. can either type travel regardless of the presence of a medium?
Dmitry_Shevchenko [17]

Yes, electromagnetic can travel without medium.

Mechanical waves and electromagnetic waves are two important ways that energy is transported in the world around us.

Waves in water and sound waves in air are two examples of mechanical waves.

Mechanical waves are caused by a disturbance or vibration in matter, whether solid, gas, liquid, or plasma.

Matter that waves are traveling through is called a medium.

These mechanical waves travel through a medium by causing the molecules to bump into each other, like falling dominoes transferring energy from one to the next.

Sound waves cannot travel in the vacuum of space because there is no medium to transmit these mechanical waves.

On the other hand electromagnetic waves don't require medium for its propagation.

An easy example would be light which is an EM wave reaches earth even though space has no medium.

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