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Molodets [167]
4 years ago
9

Please don’t comment unless you for sure know you’re right

Physics
1 answer:
aleksandrvk [35]4 years ago
5 0

As the gas is cooled, it will make a phase transition from gas into liquid, and upon further cooling from liquid to solid (ie. freezing). Some gases, such as carbon dioxide, skip the liquid phase altogether and go directly from gas to solid.

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
A runner drank a lot of water during a race. What is the expected path of the extra filtered water molecules?
Naddika [18.5K]

Answer:

Afferent arteriole, glomerulus, nephron tubule, collecting duct

Explanation:

Blood enters the kidney through the renal artery, a thick branch from the descending aorta. In the hilum, it is divided into several branches that are distributed through the lobes of the kidney and are branching forming numerous afferent arterioles that form the glomerular clew. It is precisely the walls of these capillaries that act as ultrafilters, allowing small particles to pass through.

Blood that flows through the <u>afferent arteriole</u> circulates through the capillary vessels of the kidney (the true capillaries that provide the kidney with oxygen and nutrients necessary for its function). These capillaries are grouped together to form the renal vein which, in turn, pours into the inferior vena cava.

Given the function of the kidneys to eliminate waste products through urine, it is not surprising that these organs are the ones that receive the most blood per gram of weight. One way to express renal blood flow is by considering the renal fraction or fraction of cardiac output that passes through the kidneys.

The regulation of blood flow in the glomeruli is achieved by three formations: the polar bearing, the Goormaghtigh cells and the dense macula. The polar bearing consists of a thickening of the afferent arteriole wall before it enters the <u>renal glomerulus</u>. The arteriole loses its elastic membrane, the endothelium becomes discontinuous and the middle tunic is arranged in two layers, formed by secretory cells: these secretory cells produce Angiotensin and Erythropoietin.

Goormaghtigh cells are arranged at an angle between afferent and effector arterioles and meet in small columns. They are closely related to polar bearing cells. Between both formations is the dense macula (or Zimmerman's dense macula) that is in contact with the distal tubule and afferent arteriole just before it penetrates the glomerulus. These three formations, polar bearing, Goormaghtigh cells and dense macula form the juxtaglomerular apparatus that regulates the blood flow in the glomerulus.

<u>Nephrons</u> regulate water and soluble matter (especially Electrolytes) in the body, by first filtering the blood under pressure, and then reabsorbing some necessary fluid and molecules back into the blood while secreting other unnecessary molecules.

The reabsorption and secretion are achieved with the mechanisms of Cotransporte and Contratransporte established in the nephrons and associated collection ducts. Blood filtration occurs in the glomerulus, a capping of capillaries that is inside a Bowman's capsule.

Liquid flows from the nephron in the <u>collecting duct</u> system. This segment of the nephron is crucial to the process of water conservation by the body. In the presence of the antidiuretic hormone (ADH; also called vasopressin), these ducts become water permeable and facilitate their reabsorption, thus concentrating the urine and reducing its volume. Conversely, when the body must remove excess water, for example after drinking excess fluid, ADH production is decreased and the collecting tubule becomes less permeable to water, making the urine diluted and abundant.

6 0
3 years ago
10. Inertia causes your books on the car seat to continue moving forward even after you brake. TrueFalse 11. Tires with little o
Charra [1.4K]

10. True

11. True

12. True

13. False

14. False

Hope this helps!!

3 0
3 years ago
A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1
zubka84 [21]

Explanation:

(1).  Formula to calculate the potential difference is as follows.

       \Delta V = -\int E dr

                  = -\int \frac{kq}{r^{2}} dr

                 = \frac{kq}{r_{f}} - \frac{kq}{r_{i}}

                 = \frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}

                 = \frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}

                = 38.7 volts

Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2).  Now, formula to calculate the energy stored in the capacitor is as follows.

           E = \frac{1}{2}QV

              = \frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V

              = 6.39 \times 10^{-8} J

Thus, the electric-field energy stored in the capacitor is 6.39 \times 10^{-8} J.

7 0
3 years ago
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Which of these is true?
Genrish500 [490]
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4 0
4 years ago
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