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kap26 [50]
2 years ago
12

Why is blood in the right atrium low in oxygen

Physics
1 answer:
frez [133]2 years ago
4 0

Answer:

it take blood to the lung so the blood could be oxygenated

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
1 pts
mezya [45]

Answer:

46 dgres is the answer

Explanation:

<em>plz</em><em> </em><em>mark</em><em> </em><em>me</em><em> brainliest</em><em> plz</em><em> Plz</em>

4 0
2 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
Unit 5 activity 2 I need answer
saul85 [17]

Answer:

Can you explain more in detail what it is?

Explanation:

6 0
3 years ago
Two lightbulbs have powers P1 and P2 when separately connected across an ideal battery with potential difference ΔV. The bulbs a
jarptica [38.1K]

Answer:

P'=\dfrac{P_1P_2}{P_1+P_2}

Explanation:

Lets take

Resistance of bulb 1 =R₁

Resistance of bulb 2 =R₂

As we know that power P

P= ΔV²/R

Given that voltage difference is same for both bulbs

So

P₁R₁= ΔV²       --------1

P₂R₂= ΔV²   -----------2

When these resistance are connected in series then equivalent resistance R

R=R₁+ R₂

The new power P'

P'=ΔV²/R

P'R=ΔV²   ------3

From equation 1 ,2 and 3

P'(R₁+ R₂) = ΔV²

P'\left(\dfrac{\Delta V^2}{P_1}+\dfrac{\Delta V^2}{P_2}\right)=\Delta V^2

P'=\dfrac{P_1P_2}{P_1+P_2}

6 0
3 years ago
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