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liberstina [14]

2 years ago

14

The peak wavelength of light coming from a giant red star is 660 nm. Calculate the approximate surface temperature of this star

in degrees Fahrenheit.

1 answer:

laiz [17]2 years ago

5 0

Answer:

The value is $T =260.33 \ F$

Explanation:

From the question we are told that

The the peak wavelength is $\lambda_p = 660 nm = 660 *10^{-9} \ m$

Generally according to the Wien's displacement law

$\lambda_p * T = 2.898*10^{-3} \ m \cdot K$

Here T is the approximate surface temperature of this star in K so

$660*10^{-9} * T = 2.898*10^{-3} \ m \cdot K$

=> $T = 4091 \ K$

Converting to Fahrenheit ,

$T = [400 - 273.15 ] * \frac{9}{5} + 32$

=> $T =260.33 \ F$

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Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2

posledela

A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.

8 0

3 years ago

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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago

Define any two characteristics of sound

Oliga [24]

Answer:

five characteristics: Wavelength, Amplitude, Time-Period, Frequency and Velocity or Speed

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3 years ago

What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?

GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

$E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}$

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

$E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}$

E(0.89) = 306500 N/C

3 0

3 years ago

Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

solong [7]

Answer:

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

Explanation:

For laminar flow between two parallel horizontal plates, the volumetric flow per metre of width is given as

q = (2h³/3μ) (ΔP/L)

h = hydraulic depth = 4mm/2 = 2mm = 0.002 m

μ = viscosity of oil (SAE 30) at 15.6°C = 0.38 Pa.s

(ΔP/L) = 26 KPa/m = 26000 Pa/m

q = (2h³/3μ) (ΔP/L)

q = (26000) × (2(0.002³)/(3×0.38))

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

5 0

3 years ago

Read 2 more answers