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o-na [289]
3 years ago
14

The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo

ving and is it moving towards the Earth or away from the Earth?
1- 904.3 km/s away from the Earth
2- 904.3 km/s towards the Earth
3- 907.0 km/s towards the Earth
4- 907.0 km/s away from the Earth
Physics
1 answer:
agasfer [191]3 years ago
3 0

As we know by Doppler's effect of light we have

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

3.01 \times 10^{-3} = \frac{v}{c}

here we have

v = 904.3 km/s

since wavelength is increased so we can say that it is moving away

so correct answer is

1- 904.3 km/s away from the Earth

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
2 years ago
The average Intensity of the electromagnetic field due to sunlight at the surface of the earth is 1400 W/m^2. Calculate the maxi
Ann [662]

Answer:

1027 N/C

3.42 x 10⁻⁶ T

Explanation:

I = Intensity of electromagnetic field = 1400 W/m²

E₀ = Maximum value of electric field

Intensity of electromagnetic field is given as

I = (0.5) ε₀ E₀² c

1400 =  (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²

E₀ = 1027 N/C

B₀ = maximum value of magnetic field

using the equation

E₀ = B₀ c

1027 = B₀ (3 x 10⁸)

B₀ = 3.42 x 10⁻⁶ T

4 0
3 years ago
one of the ways, covered in the article and in class, that we charged an object was rubbing it against carpet on the floor true
astraxan [27]

Answer:

It is possible to statically charge objects by rubbing it against carpet fibers, but I'm not sure if that was in the article that you read.

Explanation:

Static charge can build up via carpet fibers.

5 0
2 years ago
Can someone help with this???
Rus_ich [418]

no BECQUSE POSUM BROOB SHSHSJ

8 0
3 years ago
Read 2 more answers
Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state
topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

<u>Given  </u>

<u><em>Process 1 ---> 2 </em></u>

The relation of the process P = constant

Pressure of point (1) P1 =  10 bar = P2

Volume of point (1) V1   = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant  

<u>Process 2 ---> 3 </u>

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

<u>Process 3 ---> 1 </u>

The relation of the process PV = constant  

<u>Required  </u>

Sketch the processes on the PV coordinates

The work for each process in kJ  

<u>Solution  </u>

The work is defined by  

W=\int\limits^a_b {x} \, dx

<em>a=V2</em>

<em>b=V1</em>

<em>x=P</em>

<em>dx=dV</em>

<u>Process 1 ---> 2  </u>

P3 = P4 = 5 bar  

W=\int\limits^a_b {x} \, dx

<em>a=V3</em>

<em>b=V2</em>

<em>x=4</em>

<em>dx=dV</em>

putting the value of a, b, x, dx in above integral

W=400 kJ

<u>Process 2 ---> 3 </u>

V = constant Then there is no change in the volume,hence W = 0 kJ  

<u>Process 3 ---> 1  </u>

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1  

 W=\int\limits^a_b {x} \, dx

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

  = -160.944 KJ

5 0
3 years ago
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