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TEA [102]
2 years ago
9

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

7 m. What wavelength would be observed on Earth from that hydrogen atom?
Physics
1 answer:
Mumz [18]2 years ago
8 0

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

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A 2 kg soccer ball is traveling 28.62m/s when it hits the wall and bounces off of the wall with a velocity of 20 m/s. If the wal
Lesechka [4]

Answer:

0.149 s or 0.15 s

Explanation:

let initially ball is moving towards left  hence initial velocity = - 28.62 m/s

final velocity as ball moves right = +20 m/s

force  = rate of change in momentum

force  = mass × change in velocity / time

or time = mass × change in velocity / force

    time = 2× ( 20 -( -28.62)) / 652.36

    time  = 2× ( 20  +28.62)) / 652.36  

time  =   2× 48 .62/652.36

 time =  0.149 s or 0.15 s

5 0
3 years ago
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert
WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h

\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h

h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g} (1)

Where:

h - Maximum height of the wood block, in meters.

v - Initial speed of the block, in meters per second.

\mu_{k} - Kinetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

m - Mass, in kilograms.

s - Distance travelled by the wood block along the wooden ramp, in meters.

\theta - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that v = 10\,\frac{m}{s}, \mu_{k} = 0.20 and g = 9.807\,\frac{m}{s^{2}}, then the height reached by the block above its starting point is:

h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

h = 4.249\,m

The wood block reaches a height of 4.249 meters above its starting point.

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3 years ago
How many cm3 are there in 1 dm3?
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The answer is in the picture.

8 0
2 years ago
A 300N box on a 43 degree angle.
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Answer:is this a question??? I’m so confused

Explanation:

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2 years ago
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Yakvenalex [24]

Answer:

The hunter should aim directly at the perched monkey because the tranquilizer dart will fall away from the line sight at the same rate that the monkey falls from its perch.

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The time for the monkey to reach the ground is

t = (2 h / g)^1/2 = (18 / 9.8)^1/2 = 1.36 sec

So the horizontal speed of the dart must be at least

Vx = 90 m / 1.36 sec = 66.4 m/s

Vx = V cos theta

V = 66.4 m/s / cos 5.71 = 66.7 m/s

7 0
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