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2 years ago

What is the similarity and the principal difference between a beam of X-rays and a beam of light?

Physics

1 answer:

kati45 [8]2 years ago

3 0

Explanation:

Both are part of a family in physics which scientists refer to as the electromagnetic spectrum. All waves in this spectrum including light and x-rays travel at the same speed (3*10^8 m/s in a vacuum). However they differ in their wavelength and frequencies through the equation-----speed of light=wavelength multiplied by frequency.

Light may have a higher frequency than x-rays but lower wavelength. Either way they both multiply to receive the speed of light as the answer.

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What is the gravitational potential energy of a 0.1 kg apple sitting on top of a 12.5 m flagpole? Note: GPE= mgh. m= mass, g= ac

Pachacha [2.7K]

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2 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0

3 years ago

Rank the following in terms of increasing momentum:

professor190 [17]

It’s b because if you’re running at 5 miles per second being a 100kg weight is the fastest

5 0

3 years ago

Read 2 more answers

If mechanical energy is conserved in a system the energy at any point in time can be in the form of

I am Lyosha [343]

potential, kinetic, elastc energies

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2 years ago

A boy throws a ball vertically up. It returns to the

NikAS [45]

Answer:

31.25 m

25m/sec

Explanation:

Given :-

Time = 5sec

V = 0 (in going up)

U = 0 (in comming down)

Find :-

H and U by which it is thrown up

Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .

We know that ,

V = U + gt

0 = U - 10*2.5

U = 25 m/sec

Also,

V² = U² +2gs

0 = 625 - 20s

s = 625/20 = 31.25 m

7 0

2 years ago