Answer:
A blackbody, or Planckian radiator, is a cavity within a heated material from which heat cannot escape. No matter what the material, the walls of the cavity exhibit a characteristic spectral emission, which is a function of its temperature.
Example:
Emission from a blackbody is temperature dependent and at high temperature, a blackbody will emit a spectrum of photon energies that span the visible range, and therefore it will appear white. The Sun is an example of a high-temperature blackbody.
Questions Diagram is attached below
Answer:
Explanation:
From the question we are told that:
Speed of Train
Angle
Height of window
Width of window
The Horizontal distance between B and A from Trigonometric Laws is mathematically given by
Therefore
Distance from A-A
Therefore
Time Required to travel trough d is mathematically given as
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
Answer:
Explanation:
Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .
After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4
After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .
So,
1 / 8 the of initial light passes through the last filter .