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laiz [17]
3 years ago
10

Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b

e warmer, and why?
Physics
1 answer:
Ainat [17]3 years ago
4 0

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

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Write the advantage of dry cell over a simple cell​
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Answer:

Dry cells have high energy density or power stored to weight ratio.

Explanation:

(1) dry cell are light in weight and small in size

(2) dry cell can be transported from one place to another

(3) there is no fear of leakage in dry cell

(4) they can easily be used to run simple electrical devices

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3 years ago
Which of the following statements is true? Sound waves create areas of high and low pressure. Areas of high pressure are called
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All of the above

Explanation:

Sound waves are mechanical waves consisting of the oscillations of the particles in a medium. They are longitudinal waves, which means that the vibrations of the particles occur in a direction parallel to the direction of propagation of the wave.

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3 years ago
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Which energy transformation occurs in an
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3 years ago
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If Noah runs three laps around a 200 meter track in two minutes, what is the average speed of Noah in m/s?
devlian [24]
• 3 laps of 200 meter track = 600 meters
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3 years ago
The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are
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Answer with explanation:

Part a)

v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s

Part b)

a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}

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