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natka813 [3]
3 years ago
11

List six forms of electromagnetic radiation from the shortest waves(highest energy) to the longest waves (lowest energy)

Physics
1 answer:
romanna [79]3 years ago
8 0
That's the reverse of RIVUXG
so your answer is
gamma rays
x rays
ultraviolet light
visible light
infrared
radio waves
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The amount of current will increase since they are inversely proportional
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<h3>Article's credibility</h3>

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2 years ago
Why is water sometimes called the universal solvent.
Mrrafil [7]

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2 years ago
Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary
balandron [24]
A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
W=mg
where
m is the mass
g= \frac{GM}{r^2} is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, g=9.81 m/s^2, so Gary's weight is
W=mg=9.81 m  (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is 
R=6370 km
So the distance of Gary from the Earth's center is
r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m

The Earth's mass is M=5.97 \cdot 10^{24} kg, so the gravitational acceleration is
g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2

Therefore, Gary's weight at 300 km above Earth's surface is 
W' = mg' = 8.95 m (2)

If we compare (1) and (2), we find that Gary's weight has changed by
\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.
6 0
3 years ago
A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N i
DaniilM [7]

Answer:

The apparent weight of the person as she pass the highest point is  N  =  458.8 \ N

Explanation:

From the question we are told that

   The radius of the Ferris wheel is r = 5.0 \ m

    The period of revolution is T = 8.0 \ s

     The weight of the person is  W  =  670 \ N

   

Generally the speed of the wheel is mathematically represented as

      v =  \frac{2 \pi r}{T }

substituting values

      v =  \frac{2 * 3.142 *  5}{8 }

       v =  3.9 3 \ m/s

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

          N  =  mg  - \frac{mv^2}{r}

Where m is the mass of the person which is mathematically evaluated as

     m =  \frac{W}{g}

substituting values

    m =  \frac{670}{9.8}

    m =  68.37 \ kg

So

    N  =  68.37 * 9.8   - \frac{68.37 * {3.93}^2}{5}

    N  =  458.8 \ N

5 0
3 years ago
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