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Viktor [21]
2 years ago
5

A star is measured to be 18.9 light years from earth. what is the correct answer in the calculation shown below? (consider both

significant digits and scientific notation.)
Physics
2 answers:
Lapatulllka [165]2 years ago
4 0
A light year is a measurement of distance that is used in the space. It is the distance that a light traveled for 1 year on Earth or 365 days on Earth at a speed of 3 x 10^8 meters per second. To convert the units from light year to the SI unit meters, we simply make use of the definition of the unit. We do as follows:

18.9 light years ( 365 days / 1 light year ) ( 24 h / 1 day ) ( 3600 s / 1h ) ( 3 x 10^8 m/s) = 1.79 x 10^17 meters

Therefore, 18.9 light years is equal to 1.79 x 10^17 meters.
Temka [501]2 years ago
3 0

Answer:

1.79 ⋅ 10^14

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A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

3 0
2 years ago
When light hits the boundary between two different materials, it can undergo when light hits the boundary between two different
Rashid [163]
When light hits the boundary between two different materials, it can undergo both reflection and refraction.

Reflection is the change in the direction of the wave that strikes the boundary between two materials.<span> It involves a change in the direction of waves when they clash with an obstacle.


Refraction  involves the change in the direction of waves as they move from one medium to  </span><span><span>another followed</span></span><span> by a change in speed and wavelength (this second medium should have different permitivity for the light to change its initial properties.)</span>




3 0
2 years ago
Read 2 more answers
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
2 years ago
A green laser pointer has a wavelength of 532 nm. what is the energy of one mol of photons generated from this device?
PSYCHO15rus [73]

We have energy E = hc/λ, where h is Planck's constant c is speed of light and λ is the wavelength.

So Energy , E=\frac{6.63*10^{-34}*3*10^8}{532*10^{-9}} =3.73*10^{-19}J

Energy of one mol = 3.73*10^{-19}*6.023*10^{23}=225 kJ/mol

Energy of one mol of photons generated from this device = 225 kJ

3 0
2 years ago
A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu
faltersainse [42]
<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>
8 0
2 years ago
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