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Dmitrij [34]
3 years ago
11

A blackbody curve relates _____. intensity of radiation to energy intensity of radiation to wavelength wavelength to energy surf

ace area to wavelength
Physics
2 answers:
storchak [24]3 years ago
6 0
Intensity of radiation to wavelength
neonofarm [45]3 years ago
5 0

A blackbody curve represents the relation between <u>intensity of radiation with wavelength.</u>

Here in this curve we can see that all ideal blackbody radiates almost all wavelength of radiations and these radiations are of different intensity.

here intensity will be maximum for a given wavelength of radiation and the relation of this wavelength with the temperature of the object is given by Wein's law

It is given by

\lambda = \frac{b}{T}

now if we increase the temperature the maximum intensity for which wavelength is given will shift to the left.

Using this all we can also compare the temperature of two blackbody for which radiation graph is given to us.

You might be interested in
How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav
JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

so maximum height on moon is 6 times more than the maximum height on Earth

8 0
3 years ago
A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i
anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

  • It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

       I = \frac{1}{2}*m*r^{2}  = \frac{1}{2}*0.400 kg*(0.0865m)^{2}  = 1.5e-3 kg*m2

3 0
3 years ago
How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example
MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0
3 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
3 years ago
The s.I unit of R so that the equation velocity= R × density S dimensionally correct R is a constant
Leviafan [203]

Answer:

R = m⁴/kg . s

Explanation:

In this case, the best way to solve this is working with the units in the expression.

The units of velocity (V) are m/s

The units of density (d) are kg/m³

And R is a constant

If the expression is:

V = R * d

Replacing the units and solving for R we have

m/s = kg/m³ * R

m * m³ / s = kg * R

R = m * m³ / kg . s

<h2>R = m⁴ / kg . s</h2>

This should be the units of R

Hope this helps

5 0
3 years ago
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