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2 years ago

The force between two charges, q, and 92, is F. If the distance between the

Physics

1 answer:

Sholpan [36]2 years ago

4 0

Answer:

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

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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−

Mumz [18]

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.

8 0

2 years ago

how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance

Sergio039 [100]

Answer:

W = F * s

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g if applied force is in the horizontal direction

8 0

2 years ago

A student observes a shadow move across the sun during a solar eclipse. Which list represents the position of Earth, the sun, an

Liono4ka [1.6K]

It goes sun moon earth the moon is blocking us from seeing the sun.

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3 years ago

Why is it wrong to leave our light on

Dennis_Churaev [7]

Answer:

you will get huge electricity bills ............

8 0

2 years ago

Nancys airplane trip took 4 hour. For one-fourth of that time, the airplane flew at a speed of 880 km/h, it flew at a speed of 6

9966 [12]

So for a minute lets ignore the 880 km/h. If it took 4 hours and she flew at 600 km/h 600*4=2400. Now lets Look at the 880 bit. If it took 4 hours and she where to fly at 880 it would've been 880*4=3520. Lets do 2400-600=1800, now we've got the 600 kmh bit done. Now lets see if you fly 880 km/h for one hour then you add 1800+880=2680.

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4 0

2 years ago