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Viktor [21]
3 years ago
5

The blackbody radation emmitted from a furnace peaks at a wavelength of 1.9 x 10^-6 m (0.0000019 m). what is the temperature ins

ide the furnace?
Physics
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

Temperature, T = 1542.10 K

Explanation:

It is given that,

The black body radiation emitted from a furnace peaks at a wavelength of, \lambda=1.9\times 10^{-6}\ m

We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.

\lambda\propto \dfrac{1}{T}

or

\lambda=\dfrac{b}{T}

b = Wein's displacement constant

\lambda=\dfrac{2.93\times 10^{-3}}{T}

T=\dfrac{2.93\times 10^{-3}}{\lambda}

T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}

T = 1542.10 K

So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.

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3 years ago
a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
8 0
2 years ago
What is the effect of pressure on the solubility of gases in liquids?
34kurt
The solubility of gases in liquids increases with the increase in pressure.
3 0
2 years ago
A fan cart initially has an acceleration of 1.6m/s/s when it's fan is directed straight backwards. If you rotate the fan by 45°,
Sholpan [36]

Answer:

A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)

a. 45%

b. 29%

c. 71%

d. 50%

The correct answer is d.

d. 50%

Explanation:

Fan cart acceleration = 1.6 m/s²

Thrust = 0.25×π×D²×ρ×v×Δv

where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D

or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv

=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)

That is the thrust reduces by 50 %

3 0
3 years ago
Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?
Licemer1 [7]
Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

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Foot-pounds are gross units. Better to work in SI units when you can!

8 0
3 years ago
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