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svet-max [94.6K]
3 years ago
12

What radiation lies at frequencies just below the frequencies of visible light?

Physics
1 answer:
vlabodo [156]3 years ago
3 0
<span>Infrared radiation also called as infrared light was discovered by Sir William Herschel in 1800.It is an electronic magnetic radiation with longer wavelengths than those of a visible light.It involves waves rather than particles. It lies at frequencies just below the frequencies of visible light.</span>
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A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
emmasim [6.3K]

• Net vertical force on the block:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

(<em>n</em> = magnitude of normal force, <em>w</em> = weight)

<em>n</em> = <em>w</em> = <em>m g</em>

(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)

<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ <em>F</em> = -<em>f</em> = <em>m a</em>

(<em>f</em> = mag. of friction, <em>a</em> = acceleration)

We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) <em>a</em>

<em>a</em> = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>

∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0
2 years ago
Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just supp
xz_007 [3.2K]
The rock will continue to travel in a straight line with a constant velocity for ever... The reason is, once it leaves your hand there is no force acting on the rock, so it will just continue to move in a natural motion which is constant velocity.
6 0
2 years ago
Hii please help i’ll give brainliest if you give a correct answer please
Diano4ka-milaya [45]

Answer:

A

Explanation:

6 0
2 years ago
Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 mill
luda_lava [24]

Answer:

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

Explanation:

The distance of Earth from the Sun is 149.6\times 10^{9}\,m and of Mars from the Sun is 227.9\times 10^{9}\,m. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:

a_{r} = \frac{v^{2}}{R}

Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:

v=\frac{2\pi\cdot r}{\Delta t}

Earth:

v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{earth}=29806.079\,\frac{m}{s}

Mars:

v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{mars}=24124.244\,\frac{m}{s}

Now, centripetal accelarations can be found:

Earth:

a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}

a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}

Mars:

a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}

a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:

\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

7 0
3 years ago
Why do you think modern America needs so much energy to function?
MariettaO [177]

Answer:

electricity consumption in United states was about 3.9 trillion kilolwatthours(kWh) in 2019 electricity is an essential part of morden life

3 0
3 years ago
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