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.
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
<span>Is it true that nighttime air temperatures on a cloudy night are lower than they would be on a clear night?</span>
The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.
<h3>What current flows through the bulb as well as the power of the bulb?</h3>
From ohm's law; V = I × R
Where V is the voltage, I is the current and R is the resistance.
Also, Power is expressed as; P = V × I
Where V is voltage and I is current.
Given that;
- Resistance R = 10.0 ohms
- Voltage V = 12.0V
- Current I = ?
- Power P = ?
First, we determine the current flow through the bulb.
V = I × R
12.0V = I × 10.0 ohms
I = 12.0 ÷ 10.0
I = 1.2A
Next, we determine the power of the bulb.
P = V × I
P = 12.0V × 1.2A
P = 14.4 Watts
Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.
Learn more about Ohm's law here: brainly.com/question/12948166
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