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zheka24 [161]
3 years ago
5

Which of the following types of stars is the coolest? In Graph A, the curve peaks at 800 nm, in the red section of the visible l

ight spectrum. In Graph B, the curve peaks at 550 nm, in the green section of the visible light spectrum. In Graph C, the curve peaks at 450 nm, in the blue section of the visible light spectrum. In Graph D, the curve peaks at 300 nm, in the violet section of the visible light spectrum.
Physics
2 answers:
ASHA 777 [7]3 years ago
8 0

im no scientist but im pretty sure cool stars are warm colors and hot stars are cool colors, so red would be more cooler than blue.

Anettt [7]3 years ago
8 0

Answer:

Coolest star is corresponding to that graph which have maximum wavelength. so correct answer will be

In Graph A, the curve peaks at 800 nm, in the red section of the visible light spectrum.

Explanation:

As we know by Wein's Displacement law that

\lambda = \frac{b}{T}

here it shows that

\lambda = wavelength corresponding to maximum intensity

T = temperature

so here if wavelength and temperature is inversely depends on each other so we can say if wavelength corresponding to maximum intensity is small then its temperature must be high.

In graph A wavelength is 800 nm

In graph B wavelength is 550 nm

In graph C wavelength is 450 nm

In graph D wavelength is 300 nm

so here coolest star must be corresponding to graph A

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A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t
vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

\mu =7.20 g/m = 0.0072 kg/m is the mass linear density

Solving the equation,

f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

f=\frac{v}{2L}

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m

4 0
3 years ago
Why does the temperature decreases at higher altitudes
sergey [27]

temperature decreases at higher altitudes because as air rises the pressure decreases.

3 0
3 years ago
It is known that a shark can travel at a speed of 17 m/s. How far can a shark go in 8 seconds?
o-na [289]

Answer:136 m

Explanation:17*8

6 0
3 years ago
The fully loaded bus accelerates uniformly from rest to a speed of 14 m / s. The time taken to reach a speed of 14 m / s is 20 s
tino4ka555 [31]

Answer:

0.7m/s^2

Explanation:

acceleration=(final-initial velocity)/time

x=(14-0)/20

x=14/20

x=0.7

3 0
3 years ago
A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be​
zhuklara [117]

The question is incomplete, the complete question is;

A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?

Answer:

See explanation

Explanation:

From Newton's law of cooling;

θ1 - θ2/t = K(θ1 + θ2/2 - θo]

Where;

θ1 and θ2 are initial and final temperatures

θo is the temperature of the surroundings

K is the constant

t is the time taken

Hence;

100 - 60/5 = K(100 + 60/2 - θo)

100 - 40/10 = K(100 + 40/2 - θo)

8= (80 - θo)K -----(1)

6= (70 - θo)K -----(2)

Diving (1) by (2)

8/6 = (80 - θo)/(70 - θo)

8(70 - θo) = 6(80 - θo)

560 - 8θo = 480 - θo

560 - 480 = -θo + 8θo

80 = 7θo

θo = 11.4°

Again from Newton's law of cooling;

θ = θo + Ce^-kt

Where;

t= 0, θ = 60° and θo = 11.4°

60 = 11.4 + C e^-K(0)

60 - 11.4 = C

C= 48.6°

To obtain K

40 = 11.4 + 48.6e^-10k

40 -11.4 = 48.6e^-10k

28.6/48.6 = e^-10k

0.5585 = e^-10k

-10k = ln0.5585

k= ln0.5585/-10

K= 0.0583

Hence, the temperature in 15 minutes;

θ= 11.4 + 48.6e^(-0.0583 × 15)

θ= 31.7°

4 0
3 years ago
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