A hot-air balloon is descending at a rate of 2.3 m/s when a pas- senger drops a camera. If the camera is 41 m above the ground w
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Here , the density of the material is 4g cm³ but it is not given in CGS system.
$\sf{As\:we\:know\:that:}$
$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$
$\space$
$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$
$\space$
$\sf{It\:is\:given\:that:}$
In the system of units the mass is 100gram.
$\space$
Hence,
$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$
$\space$
In the system of units,the length is 10cm.
Henceforth,
$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$
$\space$
<u>☆</u><u> </u><u>Substitute</u><u> </u><u>the</u><u> </u><u>required</u><u> </u><u>values</u><u> </u><u>in</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>formula</u><u>-</u>
$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$
$\space$
$\sf\underline\bold{Density\:of\:the\:material:}$
= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$
$\space$
= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$
$\space$
= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$
$\space$
$\sf\underline\bold\blue{=40\:units}$
$\sf\small{Therefore,option\:2nd\:is\:correct!}$
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The speed of light and the propagation of errors allows to find the results on the questions of the radiation emitted by the laser are:
a) The frequency is: f = 3.7 10¹⁴ Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
a) The speed of a wave is related to its wavelength and frequency.
c = λ f
Where c is the speed of light, λ the wavelength and f the frequency.
They indicate that the wavelength is λ = 800 nm = 800 10⁻⁹ m, the speed of light is a constant c = 2.99 10⁸ m/s.
f =
F = 3.7 10¹⁴ Hz
b) Planck's equation states that the energy is proportional to the frequency of the radiation.
E = h f
Where E is the energy, h the Planck constant and f the frequency.
E = 6.63 10⁻³⁴ 3.7 10¹⁴
E = 2.46467 10⁻¹⁹ J
The uncertainty or error is the fluctuation that a magnitude may have due to the precision in the measurements, when the magnitude is calculated by some formula, the propagation of these uncertainties must be carried out.
Δm = ∑
the expression for energy is:
E =
When the error in the measured magnitude is not explicitly indicated, we assume that the error is in the last digit written, therefore
Δλ = ± 1 nm = ± 1 10⁻⁹ m
We calculate.
ΔE = 3.1 10⁻²² J
the error is given with a significant figure.
ΔE = 3 10⁻²² J = 0.004 10⁻¹⁹ J
The result of the energy is:
E = (2.465 ± 0.004) 10⁻¹⁹ J
In conclusion, using the speed of light and the propagation of errors, we can find the results on the questions of the radiation emitted by the laser are:
a) The frequency is; f = 3.7 1014 Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
Learn more here: brainly.com/question/15557220