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3 years ago

From class, weight is a measurement of the force of

Physics

1 answer:

Ahat [919]3 years ago

7 0

Answer:

the answer is well known as gravity

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1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out i

Reptile [31]

Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

Acceleration = -12/10.5

Acceleration = -1.14 m/s^2

Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

8 0

3 years ago

A soap bubble, when illuminated with light of frequency 5.27 Hz × 1014 Hz, appears to be especially reflective. If it is surroun

denis23 [38]

Answer:

$1.07004\times 10^{-7}\ m$

Explanation:

$n_s$ = Refractive index of bubble = 1.33

f = Frequency of light = $5.27\times 10^{14}\ Hz$

c = Speed of light = $3\times 10^8\ m/s$

The wavelength of light is given by

$\lambda=\dfrac{2n_st}{m-\dfrac{1}{2}}$

Wavelength is also given by

$\lambda=\dfrac{c}{f}$

m = 1 for minimum thickness

$\dfrac{c}{f}=\dfrac{2n_st}{m-\dfrac{1}{2}}\\\Rightarrow t=\dfrac{m-\dfrac{1}{2}c}{2n_sf}\\\Rightarrow t=\dfrac{(1-\dfrac{1}{2})\times 3\times 10^8}{2\times 1.33\times 5.27\times 10^{14}}\\\Rightarrow t=1.07004\times 10^{-7}\ m$

The minimum thickness is $1.07004\times 10^{-7}\ m$

4 0

3 years ago

What happens when an electron moving from the 3rd energy level to the 1st energy level?

Sholpan [36]

Answer:

A photon of wavelength 103 nm is released

Explanation:

When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.

For example, if we are talking about a hydrogen atom, the energy of the levels are:

$E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV$

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

$\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV$

In Joules,

$\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J$

We can also find the wavelength of this photon, using the equation:

$\Delta E = \frac{hc}{\lambda}\rightarrow \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm$