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nalin [4]
3 years ago
6

As the resistance increased under 25 V, the current blank. Compared to calculated currents, experimental currents proved to be b

lank . Currents varied proportionally according to resistance: 10 Ω of resistance produced a current of blank A, while 100 Ω of resistance produced a current of blank A.
Physics
1 answer:
sukhopar [10]3 years ago
4 0

Answer:

Decreased, The Same, 2.5 and .25

Explanation:

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When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?
Oduvanchick [21]
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
a = r \times ( \frac{1}{n1}  -  \frac{1}{n2} )
We insert the values

a = 2.18 \times  {10}^{ - 18}  \times ( \frac{1}{ {1}^{2} } -  \frac{1}{ {6}^{2} }  )
= 2.12 \times {10}^{ - 18}
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
wavelength =  \frac{h \times c}{a}
Finally we insert the values
\frac{6.626 \times  {10}^{ - 34}  \times 3 \times  {10}^{8} }{2.12 \times  {10}^{ - 18} }  = 9.376 \times  {10}^{ - 8}
Which is the same as 93.8 nm
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3 years ago
What’s the name of the “super continent” that broke up about 225 million years ago?
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In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear
Kamila [148]

Answer:

a)   m = 10  and    b)  λ  = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation      d sin θ = m λ

Diffraction equation         a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n  are an integer.

Let's find the relationship of these two equations

    d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

    d / a = m

Let's calculate

    m = 0.2 / 0.02

    m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

    θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

    d sin  θ = m λ

    λ = d sin θ / m

    λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

    λ  = 3.119 10⁻⁷ m

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