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3 years ago

What is Stefan Boltzmann's law??? Need explaination

Physics

2 answers:

sukhopar [10]3 years ago

7 0

Answer:

The Stefan–Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body in unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body's

Explanation:

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pav-90 [236]3 years ago

6 0

The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time {\displaystyle j^{\star }} j^{\star} (also known as the black-body radiant emittance) is directly proportional to the fourth power of the black body's thermodynamic temperature T:

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a sound has a frequency of 230 hz. what is the velocity of this sound if it has a wavelength of 46 cm?

lapo4ka [179]

Answer:

V=f×λ

Where,

V is the velocity of the wave measure using m/s.

f is the frequency of the wave measured using Hz.

λ is the wavelength of the wave measured using m.

transform 46 cm in m = 0.46m and there u have it

Explanation:

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3 years ago

ASAP What is elastic potential energy?

chubhunter [2.5K]

Answer:

Elastic potential energy is stored when materials stretch or compress. Examples of it include springs, rubber bands, and slingshots.

Explanation:

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3 years ago

A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .

Ray Of Light [21]

A) Net force in component form: $F=100.0i-300.0j$

B) Magnitude of the net force: 316.2, direction: $-71.6^{\circ}$

Explanation:

The three forces given in this problem are:

$F_1=300i+500j$

$F_2=-200i$

$F_3=-800 j$

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

$F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100$

- Along the y-direction:

$F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300$

So, the net force in component form is

$F=100.0i-300.0j$

The magnitude of a vector F is given by Pythagorean's theorem:

$|F|=\sqrt{F_x^2+F_y^2}$

where in this problem,

$F_x=100$ is the x-component

$F_y=-300$ is the y-component

Substituting,

$|F|=\sqrt{(100)^2+(-300)^2}=316.2$

The direction instead is given by

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}$

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

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8 0

3 years ago

Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th

bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

the mass of planet B to be, Mₓ = M

the radius of planet A to be, R₁

the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁² - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂² [where, R₂ = 2R₁]

= GM/4R₁² -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

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2 years ago

An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over

Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

Em₀ = U₁ + U₂

Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

y₁ = 0

y₂ = y

Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

$E_{mf}$ = K₁ + U₁ + K₂ + U₂

$E_{mf}$ = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g $y_{f}$ + m₂ g $y_{f}$

Since the masses are joined by a rope, they must have the same speed

$E_{mf}$ = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g $y_{f}$

$E_{mf}$ = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

How energy is conserved

Em₀ = $E_{mf}$

2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

3/2 v₁² = 2 g y -3/2 g y

3/2 v₁² = ½ g y

V₁ = √ (gy / 3)

5 0

3 years ago