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trasher [3.6K]
3 years ago
13

What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?

Physics
1 answer:
blsea [12.9K]3 years ago
7 0

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

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A car, starting from the origin, travels
sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0
3 years ago
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An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu
Doss [256]

Answer:

400000\ \text{N/C}

Explanation:

q_1 = Charge at 3000 m = 40 C

q_2 = Charge at 1000 m = -40 C

r_1 = 3000 m

r_2 = 1000 m

k = Coulomb constant = 9\times10^9\ \text{Nm}^2/\text{C}^2

Electric field due to the charge at 3000 m

E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}

Electric field due to the charge at 1000 m

E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}

Electric field at the aircraft is E_1+E_2=40000+360000=400000\ \text{N/C}.

7 0
3 years ago
What is a lever and mention its types<br>​
Grace [21]

Answer:

Three different types of levers exist, depending on where the input force, fulcrum, and load are. A class 1 lever has the fulcrum between the input force and load. A class 2 lever has the load between the fulcrum and input force. A class 3 lever is a lever that has the input force in between the fulcrum and the load.

Explanation:

3 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. at one position along its path it has a velocity of 26.3 m
galina1969 [7]
We will use the formula / equation to determined the time.

Distance = ½ * (vi + vf) * t 
48100 = ½ * (26.3 + 41.9) * t 
t = 48100 ÷ 34.1 = 1410.557185 seconds 

We will use the formula / equation to determined the acceleration. 

vf = vi + a * t 
41.9 = 26.3 + a * 1410.557185 
a = (41.9 – 26.3) ÷ 1410.557185 = 0.011059459 m/s^2 

We will use the formula / equation to determined the acceleration.

vf^2 = vi^2 + 2 * a * d 
41.9^1 = 26.3^2 + 2 * a * 48100 
a = (41.9^2 – 26.3^2) ÷ 96200 = 0. 011059459 m/s^2 
Since both answers are the same, I believe the acceleration is correct.
6 0
4 years ago
Ch 31 HW Exercise 31.10 7 of 15 Constants You want the current amplitude through a inductor with an inductance of 4.90 mH (part
sergey [27]

Answer:

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1) , which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and  L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz

5 0
3 years ago
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