All categories

4 years ago

What Type of relationship exists between the Temperature of a Star and the Wavelength of a Star?

2 answers:

8 0

To estimate the surface temperature of a star, we can use the known relationship between the temperature of a blackbody, and the wavelength of light were its spectrum peaks. That is, as you increase the temperature of a blackbody, the peak of its spectrum moves to shorter (bluer) wavelengths of light.

Debora [2.8K]4 years ago

3 0

This is known as Wien's Law:
The relationship is:
wavelength = 0.0029/temperature

It is an inversely proportional relationship.

You might be interested in

Assume it takes 5.00 minutes to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3).

dolphi86 [110]

   (50 gal / 5 min) x (.0037854 m³/gal) x (1 min / 60 sec)

   = (50 · 0.0037854 · 1) / (5 · 60) m³/sec

   = 0.000631 m³/sec

7 0

4 years ago

Which is less likely to be reliable source of information,the webpage of a university or the webpage of a scientist who is tryin

Nikolay [14]

The scientist because he is biased and a university isn't. The scientist will say good things about the invention to sell more of it. Hope this helps! ;)

6 0

3 years ago

Read 2 more answers

The angular position of objects as a function of time is given, where a, b, and care constants. In which of these cases is the a

GalinKa [24]

Answer:

Explanation:

The options is not well presented

This are the options

A. θ = at³ + b

B. θ = at² + bt + c

C. θ = at² — b

D. θ = Sin(at)

So, we want to prove which of the following option have a constant angular acceleration I.e. does not depend on time

Now,

Angular acceleration can be determine using.

α = d²θ / dt²

α = θ''(t)

So, second deferential of each θ(t) will give the angular acceleration

A. θ = at³ + b

dθ/dt = 3at² + 0 = 3at²

d²θ/dt² = 6at

α = d²θ/dt² = 6at

The angular acceleration here still depend on time

B. θ = at² + bt + c

dθ/dt = 2at + b + 0 = 2at + b

d²θ/dt² = 2a + 0 = 2a

α = d²θ/dt² = 2a

Then, the angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

C. θ = at² —b

dθ/dt = 2at — 0 = 2at

d²θ/dt² = 2a

α = d²θ/dt² = 2a

Same as above in B. The angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

D. θ = Sin(at)

dθ/dt = aCos(at)

d²θ/dt² = —a²Sin(at) = —a²θ

α = d²θ/dt² = -a²θ

Since θ is not a constant, then, the angular acceleration is dependent on time and angular displacement

So,

The answer is B and C

4 0

3 years ago

What particles of an atom are inside the neuculas?

Mrac [35]

*It's nucleas. There are protons, neutrons and electrons.

3 0

3 years ago

Calculate Neptune's mass given the acceleration due to gravity at the north pole is 11.529 m/s2 and the radius of Neptune at the

dolphi86 [110]

Answer:

The mass of Neptune is $1.023\times 10^{26}$ kilograms.

Explanation:

From Newton's Law of Gravitation, the gravitational acceleration of Neptune is determined by the following formula:

$g = \frac{G\cdot M}{R^{2}}$ (1)

Where:

$G$ - Gravitational constant, measured in cubic meters per kilogram-square second.

$M$ - Mass of the planet, measured in kilograms.

$R$ - Radius of the planet, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $g = 11.529\,\frac{m}{s^{2}}$ and $R = 24.340\times 10^{6}\,m$ , then the mass of Neptune is:

$M = \frac{g\cdot R^{2}}{G}$

$M = \frac{\left(11.529\,\frac{m}{s^{2}}\right)\cdot (24.340\times 10^{6}\,m)^{2} }{6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} }$

$M = 1.023\times 10^{26}\,kg$

The mass of Neptune is $1.023\times 10^{26}$ kilograms.

5 0

3 years ago