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Elden [556K]
3 years ago
9

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm(red) and 486 nm (blue). Light fr

om a hydrogen lamp illuminates a diffraction grating with 520lines/mm , and the light is observed on a screen 1.5m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
STatiana [176]3 years ago
5 0

Answer:

0.152 m

Explanation:

The condition for constructive interference is

d sinθ= mλ  (m= 0,1,2,3...)

the slit width

d= 1/N

d= 10^(-3)/520

= 1.92×10^(-6)

The angular spread for the red light is

\theta_r= sin^{-1}(\frac{m\lambda}{d})

\theta_r= sin^{-1}(\frac{656\times10^{-9}}{1.92\times10^{-6}})

= 19.97°

The angular spread of blue light is

\theta_b= sin^{-1}(\frac{m\lambda}{d})

\theta_b= sin^{-1}(\frac{456\times10^{-9}}{1.92\times10^{-6}})

=14.66°

The distance between the first order red fringe and blue fringe is,

y= y_r- y_b = Ltanθ_r -Ltanθ_b

=1.50( tan19.97°-tan14.66°) = 0.152 m

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Explain why an object’s weight is dependent upon where in the universe it is located.
strojnjashka [21]

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the  law of universal gravity

F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]

Now replacing the values in the equation we have:

F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a  radius of 71492 [km]. Find the force exerted by the planet upon the person.

F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} }  \\F= 1734.73[N]

We will calculate the value of the gravity accelaration of Jupiter. Let's check

g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

6 0
4 years ago
An old clock has a spring that must be wound to make the clock hands move. Which statement describes the energy of the spring an
topjm [15]

Answer:

c

Explanation:

neither the spring or hands are in the action of movemnet

8 0
3 years ago
Even though earth has ____ mass than the sun, the moon orbits earth because it much nearer to it.
kozerog [31]

Even though the Earth has less mass than the Sun, the moon orbits Earth because it’s much nearer to it.

<u>Explanation :</u>

The fact is that the Moon orbits both the Sun and the Earth. On looking at the orbit of the Moon, it orbits in the same manner the way Earth does, but in a Spiro graph pattern along with orbiting the Earth with a small wobble to it.  

Since the Sun has greater distance from the Moon as compared to the Earth (around 400 times), the gravity of Earth draws better impact on the Moon.  

The escape velocity of the Moon is about 1.2 km/s at the distance from the Earth which is not sufficient to get ripped away from the Earth.  

Hence, the moon orbits the Earth along with orbiting the Sun together with the Earth, but seems as if it only orbits the Moon.

3 0
3 years ago
Read 2 more answers
It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft
Paul [167]

Answer: 1.23\ m/s^2

Explanation:

Given

At an elevation of y=34.7\ km, spacecraft is dropping vertically at a speed of u=293\ m/s

Final velocity of the spacecraft is v=0

using equation of motion i.e. v^2-u^2=2as

Insert the values

\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2

Therefore, magnitude of acceleration is 1.23\ m/s^2.

8 0
3 years ago
2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan
ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

5 0
3 years ago
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