All categories

3 years ago

which of the following ideas provided an important step toward a more complete understanding of blackbody radiation?

2 answers:

7 0

The theory of blackbody radiation is shown by the equation. E= h×f
Where E = energy of the photon, f is the frequency of photon and h is equal to 6.63 ×10⁻³⁴ J.s in the plank constant. Light exists in form of a bundle of energy which is called photon and its energy is determined by frequency.
Therefore, any emission or absorption of light must involve discrete changes in energy of the system. Light is composed of discrete chunks that must be absorbed as a unit.

tiny-mole [99]3 years ago

3 0

Answer:

Light is composed of discrete chunks that must be absorbed as units

Explanation:

You might be interested in

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a

masha68 [24]

Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
$F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N$
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
$\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m$
And by using Hook's law, we can find the constant of the spring:
$k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m$

4 0

3 years ago

If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy pf 392,000 J and a kinetic ene

Viefleur [7K]

I just woke him crying crying laughing crying and laughing I’m so mad he got me blocked him lol I got a hold on her phone the answer is 0 m

5 0

3 years ago

Electromagnets are created by?

AysviL [449]

A coil of insulated wire around an iron core

7 0

3 years ago