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makkiz [27]
3 years ago
5

Please select the word from the list that best fits the definition Which type of waves from the electromagnetic spectrum are use

d in heat lamps and heat sensing devices? visible light ultraviolet rays radio waves infrared waves
Physics
2 answers:
dalvyx [7]3 years ago
4 0
Visible light ultraviolet rays radio waves infrared waves
Naddik [55]3 years ago
4 0
<h2>Right answer: infrared waves</h2>

Any body or object that has a temperature above absolute zero (0 Kelvin) emits infrared radiation (IR).

This type of radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm).

Infrared waves can be divided into:

- Near infrared or long wave infrared: it is the least sensitive to color and is easily absorbed by water.

- Medium or medium wave infrared: it is also insensitive to color and easily absorbed by water and many types of plastics and paints.

- Far infrared or short wave infrared: it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.

It should be noted that infrared rays only cause an effect if they are absorbed. When this happens they become in thermal energy (heat).

IR light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

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A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of th
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Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

half\,\,max-height = \frac{v_{yi}^2}{4\,g}

we can use this information to find the y component of the velocity at that height via the formula:

v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o

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