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2 years ago

Help, please!!!!! I just need help putting the vocab word in the box where it belongs.

Physics

1 answer:

Ber [7]2 years ago

8 0

here's the first part but for the 2nd one all I know is that the word "compression" goes on the spirals that are closer together.

hope this helps!

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Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T. (a) What

Marat540 [252]

Answer:

Explanation

Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T

A. To calculate the energy difference in the magnetic moment orientation

∆E = 2μB

For example, any electron's magnetic moment is measured to be 9.284764×10^−24 J/T

Then

μ = 9.284764 × 10^-24 J/T

∆E = 2μB

∆E = 2 × 9.284764 × 10^-24 × 0.21

∆E = 3.8996 × 10^-24 J

Then, to eV

1eV = 1.602 × 10^-19J

∆E = 3.8996 × 10^-24 J × 1eV / 1.602 × 10^-19J

∆E = 2.43 × 10^-5 eV

B. Frequency?

To determine the frequency of radiation hitch would induce the transition between the two states is,

∆E = hf

Where h is plank constant

h = 6.626 × 10-34 Js

Then, f = ∆E / h

f = 3.8996 × 10^-24 / 6.626 × 10^-34

f = 5.885 × 10^9 Hz

f ≈ 5.89 GHz

C. The wavelength of the radiation

From wave equation

v = fλ

In electromagnetic, we deal with speed of light, v = c

And the speed of light in vacuum is

c = 3 × 10^8 m/s

c = fλ

λ = c / f

λ = 3 × 10^8 / 5.885 × 10^9

λ = 0.051 m

λ = 5.1 cm

λ = 51 mm

D. It belongs to the microwave

From table

Micro waves ranges from

•Wavelength 10 to 0.01cm

Then we got λ = 5.1 cm, which is in the range.

•Frequency 3GHz to 3 Thz

Then, we got f ≈ 5.89 GHz, which is in the range

•Energy 10^-5 to 0.01 eV

We got ∆E = 2.43 × 10^-5 eV, which is in the range of the microwave

The value above is in microwave range

5 0

3 years ago

If a range hood removes contaminants at a flow rate of F liters of air per second, then the percent P of contaminants that are

podryga [215]

Explanation:

The given value of P is as follows.

P = 1.06F + 22.18, $10 \leq F \leq 70$

or, P' = 1.06

As p' is defined and non-zero. Hence, only critical points are boundary points.

For F = 10, the value of P will be calculated as follows.

P = $1.06 \times 10 + 22.18$

= 32.78

For F = 70, the value of P will be calculated as follows.

P = $1.06 \times 70 + 22.18$

= 96.38

Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.

3 0

3 years ago

A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before fal

Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

4 0

3 years ago

Read 2 more answers

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Nataly [62]

Answer:

v = 25 m / s

Explanation:

For this exercise we use the relations and kinematics

first part, the train accelerates from v₀ = 7.0 m/s to 13 m/s in a time t = 8.0 s

v = v₀ + a t

a = $\frac{v- v_o}{t}$

a = $\frac{13-7}{8}$

a = 0.75 m / s²

second part. Accelerate again for t = 16 s

v = v₁ + a t

for this interval the initial velocity is v₁ = 13 m / s

v = 13 + 0.75 16

v = 25 m / s

4 0

3 years ago

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li

babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

$y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m$

The slit separation is 0.00001266 m

3 0

3 years ago