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Liono4ka [1.6K]

3 years ago

6

Answers n = 2 energy level to the n = 4 energy level has a __________ wavelength than the photon absorbed by an electron moving

from the n = 2 to the n = 3 energy level.

1 answer:

ludmilkaskok [199]3 years ago

3 0

The answer is The Bohr Atom. I hope it helped

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An electron moving parallel to a uniform electric field increases its speed from 2.0 ×× 1077 m/sm/s to 4.0 ×× 1077 m/sm/s over a

algol [13]

Answer:

262 kN/C

Explanation:

If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:

v² = v0² + 2aΔS

Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:

(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013

16x10¹⁴ = 4x10¹⁴ + 0.026a

0.026a = 12x10¹⁴

a = 4.61x10¹⁶ m/s²

The electric force due to the electric field (E) is:

F = Eq

Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:

F = m*a

Where m is the mass, so:

E*q = m*a

The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:

E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶

E = 261,866.42 N/C

E = 262 kN/C

4 0

3 years ago

A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?

Tatiana [17]

Answer:

<u>We are given:</u>

displacement (s) = 130 m

acceleration (a) = -5 m/s²

final velocity (v) = 0 m/s [the cars 'stops' in 130 m]

initial velocity (u) = u m/s

<u>Solving for initial velocity:</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(0)² - (u)² = 2(-5)(130)

-u² = -1300

u² = 1300

u = √1300

u = 36 m/s

8 0

3 years ago

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

galina1969 [7]

Answer:

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

5 0

2 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

2 years ago

What does aerobic refer to?

Gnesinka [82]

Answer:

A) how your body uses oxygen

7 0

2 years ago

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