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Murrr4er [49]
3 years ago
8

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

500 K, a heated metal at 1500 K, human skin at 305 K, and a cryogenically cooled metal surface at 60 K. Estimate the fraction of the solar emission that is in the following spectral regions: the ultraviolet, the visible, and the infrared.
Physics
1 answer:
igomit [66]3 years ago
8 0

Answer

Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

      \lambda_{max} = 0.5 \mu_m

2) for tungsten T = 2500 K

      \lambda_{max} = \dfrac{2898}{2500}

      \lambda_{max} = 1.16 \mu_m

3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

      \lambda_{max} = \dfrac{2898}{60}

      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

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Answer:

d

Explanation:

4 0
2 years ago
FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine
Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

\tau=Fd=mg\cdot R

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

\tau=I\alpha

therefore, equating this to the above equation gives

mg\cdot R=I\alpha

solving for alpha gives

\alpha=\frac{mgR}{I}

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}\boxed{\alpha=1.47s^{-2}}

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

\theta=\frac{1}{2}\alpha t^2

Therefore, ar t = 4.2 s, the above gives

\theta=\frac{1}{2}(1.47)(4.2)^2

\boxed{\theta=12.97}

So how many revolutions is this?

To find out we just divide by 2 pi:

\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}\boxed{\#\text{rev}=2.06}

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

\omega^2=w^2_0+2\alpha(\Delta\theta)_{}

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

\omega^2=0+2(1.47)(12.97)w^2=38.12\boxed{\omega=6.17.}

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0
1 year ago
Fa car's power output is increased, its efficiency:
Inessa [10]
It’s solved by using a pretty standard formula for efficiency.

4 0
3 years ago
B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

7 0
2 years ago
10m= (5.0) + (.5)(9.8)(5.0)
Norma-Jean [14]
M=2.45 because you multiply out the equation on the right and divide by 10
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3 years ago
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