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2 years ago

7

At rest, hydrogen has a spectral line at 116 nm. if this line is observed at 107 nm for the star sirius, how fast is sirius movi

ng in km/s?

1 answer:

Citrus2011 [14]2 years ago

4 0

2.3275862×10¹²km/s fast is sirius moving in km/s.

<h3>Briefing:</h3>

Hydrogen has a spectral line at = 116nm=116×10⁻⁹m

Line is observed at = 107 nm=107×10⁻⁹m

Now, from the Hubble's law

V= $(\Delta \lambda / \lambda)$ ×C

Where,

v is the velocity

Δλ = Change in wavelength = 116 - 107= 9nm=9×10⁻⁹m

λ = Actual wavelength=116nm=116×10⁻⁹m

C is the speed of the light=3×10⁸m/s

on substituting the respective values, we get

V=(9/116)×3×10⁸=23275862.069×10⁵m/s

V=2.3275862×10¹²km/s.

<h3>What is the wavelength?</h3>

A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. Typically, in wireless systems, this length is specified in meters (m), centimeters (cm), or millimeters (mm).

To know more about Wavelength visit:

brainly.com/question/13533093

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Dominik [7]

Answer:

The center of mass for the object is $y_c = 1.063 \ m$ from the origin

Explanation:

From the question we are told that

The mass of the first object is $m_1 = 1.99 \ kg$

The position of first object with respect to origin $y_1 = 2.99 \ m$

The mass of the second object is $m_2 = 2.96 \ kg$

The position of second object with respect to origin $y_2 = 2.57 \ m$

The mass of the third object is $m_3 = 2.43 \ kg$

The position of third object with respect to origin $y_3 = 0 \ m$

The mass of the fourth object is $m_3 = 3.96 \ kg$

The position of fourth object with respect to origin $y_3 = -0.502 \ m$

Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

$y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}$

=> $y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96 + 2.43 + 3.96}$

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3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

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Answer:

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Given data

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Solution

First we need to convert the speed of truck to inches per mile

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1 mile=63360 inches

1 hour=60 minutes

so

$v=(50*\frac{63360}{60} )\\v=52800in/min$

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3 years ago