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Anestetic [448]
3 years ago
5

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth

to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 1.61THz. Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.
Physics
1 answer:
liberstina [14]3 years ago
7 0

Answer:

λ = 1.86 x 10⁻⁴ m = 186 μm

Explanation:

The relationship between the wavelength and the frequency of a wave is given by the following equation:

c = f\lambda\\\\\lambda = \frac{c}{f}

where,

λ = wavelength of infrared radiation = ?

c = speed of infrared radiation = speed of light = 3 x 10⁸ m/s

f = frequency of infrared radiation = 1.61 THz = 1.61 x 10¹² Hz

Therefore,

\lambda = \frac{3\ x\ 10^8\ m/s}{1.61\ x\ 10^{12}\ Hz}

<u>λ = 1.86 x 10⁻⁴ m = 186 μm</u>

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A 3 column table with 5 rows labeled table A. The first column is labeled voltage in volts with entries 1.0, 5.0, 10, 20, 50. Th
Volgvan

Answer:

0.25A

1.0A

2.5A

Explanation:

this is only for the calculated current column on edge

6 0
4 years ago
Read 2 more answers
Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
3 years ago
que 2. Why do we keep frequency constant instead of keeping vibrating length constam second law of vibrating string?​
ella [17]

Answer:

The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

The law can be expressed mathematically as follows;

f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

m = \dfrac{1}{ l^2} \cdot \dfrac{T}{4\cdot f^2} }

From which it can be shown that the following relation holds with the limits of error in the experiment

m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²

Explanation:

8 0
3 years ago
A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an
elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)=(2.4-x^2)\hat{i}N

Initial position of the block=x=0

Initial velocity of block=v_i=0

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy=K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0

Work energy theorem:

K_f-K_i=W

Where K_f=Final kinetic energy

K_i=Initial kinetic energy

W=Total work done

Substitute the values then we get

K_f-0=\int_{0}^{2}F(x)dx

Because work done=Force\times displacement

K_f=\int_{0}^{2}(2.4-x^2)dx

K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}

K_f=2.4(2)-\frac{8}{3}=2.13 J

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy =K=2.4x-\frac{x^3}{3}

When the kinetic energy is maximum then \frac{dK}{dx}=0

\frac{d(2.4x-\frac{x^3}{3})}{dx}=0

2.4-x^2=0

x^2=2.4

x=\pm\sqrt{2.4}

\frac{d^2K}{dx^2}=-2x

Substitute x=\sqrt{2.4}

\frac{d^2K}{dx^2}=-2\sqrt{2.4}

Substitute x=-\sqrt{2.4}

\frac{d^2K}{dx^2}=2\sqrt{2.4}>0

Hence, the kinetic energy is maximum at x=\sqrt{2.4}

Again by work energy theorem , the  maximum kinetic energy of the block between x=0 and x=2.0 m is given by

K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx

k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}

K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0
4 years ago
Hi please I need help with my mMATHEMATICS ASSIGNMENTS
alisha [4.7K]

Answer:

The answers are below

Explanation:

a)3:5

Multiply by 4

Answer =12:20

b) 11:35

Multiply by 6 on each side

Answer =66:210

2) 68% of 375=255

375-255=120

Answer is 120 people escaped malaria

3) 5*3=15

450/15=30

Answer=30

4) 1251/9=139

278:973

There are 278 girls in the school

5 0
3 years ago
Read 2 more answers
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