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igomit [66]
2 years ago
14

Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8. Note: 1 nanometer

(nm) = .0000001 centimeters (cm) 208nm 400nm 828nm 1800nm
Physics
1 answer:
kodGreya [7K]2 years ago
8 0

The peak wavelength of Betelgeuse is 828 nm

Explanation:

The relationship between surface temperature and peak wavelength of a star is given by Wien's displacement law:

\lambda=\frac{b}{T}

where

\lambda is the peak wavelength

T is the surface temperature

b=2.898\cdot 10^{-3} m\cdot K is Wien's constant

For Betelgeuse, the surface temperature is approximately

T = 3500 K

Therefore, its peak wavelength is:

\lambda=\frac{2.898\cdot 10^{-3}}{3500}=8.28\cdot 10^{-7} m = 828 nm

Learn more about wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
2 years ago
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
The lowest point in Death Valley is 85.0 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m. What is
mestny [16]

Answer:

The change in gravitational potential energy of the hiker = 2869685 J

Explanation:

Potential Energy: This is the energy possessed by a body, due to its change in position in the gravitational field. The unit of potential energy is Joules (J)

From the question,

Change in gravitational potential energy = Energy of the hiker at the top of  Mt. Whitney - Energy of the hiker at the floor of Death valley.

ΔEp = mgh₂ - mgh₁

ΔEp = mg(h₂-h₁)........................... Equation 1

Where ΔEp =  change in Potential Energy of the hiker, m = mass of the hiker, g = acceleration due to gravity, h₁ = lowest point in Death valley, h₂ = Elevation of Mt. Whitney.

Given: m = 65.0 kg, h₁ = -85 m ( because is a valley), h₂ = 4420 m,

Constant: g = 9.8 m/s²

Note: The h₁ is negative because is below sea level.

Substituting into equation 1

ΔEp = 65×9.8×[4420-(-85)]

ΔEp = 637(4420+85)

ΔEp = 637(4505)

ΔEp = 2869685

ΔEp = 2869685 J.

Thus the change in gravitational potential energy of the hiker = 2869685 J

6 0
2 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) 1\times10^6N/C

B) 2\times10^{-13}N

C) 1\times10^{-2}V

D) Toward the inner wall

E) 3\times10^{-17}J

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula E=\frac{\sigma}{k\epsilon_0}, where \epsilon_0=8.85\times10^{-12}C/Vm and in our case \sigma=10^5C/m^2 and, for air,k=1.00059, so we have:

E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C

B) The K+ ion has one elemental charge excess, so its charge is q=1.6\times10^{-19}C, and the force a charge experiments under an electric field E is given by F=qE, so we have:

F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by \Delta V=dE, so we have:

\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference \Delta V on a charge Q is W=Q\Delta V, and is equal to the kinetic energy imparted on it, so we have:

K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J

5 0
2 years ago
Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s
Black_prince [1.1K]

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

1.91J = I/2*\omega^2

where I = 2/3*m*R^2=0.23kg.m^2

Solving for ω:

\omega = \sqrt{W*2/I} =4.07rad/s

7 0
2 years ago
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