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GarryVolchara [31]

2 years ago

5

How is Momentum Calculated ??

2 answers:

krek1111 [17]2 years ago

7 0

The momentum of a body is calculated as mass × velocity of a body.

<h3 /><h3>What is momentum?</h3>

The term momentum refers to the product of mass and velocity. The unit of momentum is Kgms-1. Hence, there are two things that compose momentum and they are;

mass
velocity

Hence, the momentum of a body is calculated as mass × velocity of a body.

Learn more about momentum: brainly.com/question/4956182

LekaFEV [45]2 years ago

7 0

<h2>╔═.✵.══════════╗</h2>

How is Momentum Calculated

<h2>╚══════════.✵.═╝</h2>

Linear Momentum:

══════⊹⊱❖⊰⊹══════

→ <u>Formula</u> ⇝ p = m × v

Here, p is the linear momentum, m is the mass of the object and v is the velocity of the object.

Angular Momentum:

══════⊹⊱❖⊰⊹══════

→<u>Formula</u> ⇝ L= m × v × r

Here, L is the angular momentum, m is the mass of the object, v is the velocity of the object and r is the radius.

<h3> ☆ Hope it's help you ☆</h3>

»»————-　♡　————-««

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Light incident on a polarizer is then passed through a second polarizer. If the polarizer and the analyzer are perpendicular to

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The first polarization will restrict the plane of oscillation for the light, and the second one will block it completely as it is perpendicular to the first.

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A heavy crate applies of 1500 N on a 25-m^2 piston . The smaller piston is 1/30 the size of the larger one . what force is neede

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I don't know the name, but the opposite of gravity, if that helps!

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3 years ago

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. A long 10-cm-diameter steam pipe whose external surface temperature is 110oC passes through some open area that is not protect

Nata [24]

Answer:

$Nu = 30.311$

Explanation:

Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:

$Nu = \left\{0.6+\frac{0.387\cdot Ra_{D}^{\frac{1}{6} }}{[1+\left(\frac{0.559}{Pr} \right)^{\frac{9}{16}} ]^{\frac{8}{27} }} \right\}^{2}$ , for $Ra_{D} \le 10^{12}$ .

Where $Ra_{D}$ is the Rayleigh number associated with the cylinder.

The Rayleigh number is:

$Ra_{D} = \frac{g\cdot \beta\cdot (T_{pipe}-T_{air})\cdot D^{3}}{\nu^{2}}\cdot Pr$

By assuming that air behaves ideally, the coefficient of volume expansion is:

$\beta = \frac{1}{T}$

$\beta = \frac{1}{283.15\,K}$

$\beta = 3.532\times 10^{-3}\,\frac{1}{K}$

The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:

$\nu = 1.426\times 10^{-5}\,\frac{m^{2}}{s}$

$\mu = 1.778\times 10^{-5}\,\frac{kg}{m\cdot s}$

$k = 0.02439\,\frac{W}{m\cdot ^{\textdegree}C}$

$c_{p} = 1006\,\frac{J}{kg\cdot ^{\textdegree}C}$

The Prandtl number is:

$Pr = \frac{\mu\cdot c_{p}}{k}$

$Pr = \frac{(1.778\times 10^{-5}\,\frac{kg}{m\cdot s} )\cdot (1006\,\frac{J}{kg\cdot ^{\textdegree}C} )}{0.02439\,\frac{W}{m\cdot ^{\textdegree}C} }$

$Pr = 0.733$

Likewise, the Rayleigh number is:

$Ra_{D} = \frac{(9.807\,\frac{m}{s^{2}} )\cdot (3.532\times 10^{-3}\,\frac{1}{K} )\cdot (110^{\textdegree}C-10^{\textdegree}C)\cdot (0.1\,m)^{3}}{(1.426\times 10^{-5}\,\frac{m^{2}}{s})^{2} }\cdot (0.733)$

$Ra_{D} = 12.486\times 10^{6}$

Finally, the Nusselt number is:

$Nu = \left\{0.6+\frac{0.387\cdot (12.486\times 10^{6})^{\frac{1}{6} }}{\left[1 + \left(\frac{0.559}{0.733}\right)^{\frac{9}{16} }\right]^{\frac{8}{27} }} \right\}^{2}$

$Nu = 30.311$

8 0

3 years ago

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A light with frequency of 3.5 x 10^15 Hz is utilized to illuminate the selenium metal (work function 5.11 eV). What will be the

ruslelena [56]

Answer:

The maximum speed of the ejected photoelectrons is 1.815 x 10⁶ m/s.

Explanation:

Given;

frequency of the light, f = 3.5 x 10¹⁵ Hz

work function of the metal, Φ = 5.11 eV

Φ = 5.11 x 1.602 x 10⁻¹⁹ J = 8.186 x 10⁻¹⁹ J

The energy of the incident light is given as sum of maximum kinetic energy and work function of the metal.

E = K.E + Φ

where;

E is the energy of the incident light, calculated as;

E = hf

E = (6.626 x 10⁻³⁴)(3.5 x 10¹⁵)

E = 2.319 x 10⁻¹⁸ J

The maximum kinetic energy of the photoelectrons is calculated as;

K.E = E - Φ

K.E = 2.319 x 10⁻¹⁸ J - 8.186 x 10⁻¹⁹ J

K.E = 2.319 x 10⁻¹⁸ J - 0.8186 x 10⁻¹⁸ J

K.E = 1.5004 x 10⁻¹⁸J

The maximum speed of the ejected photoelectrons in 10⁶ m/s is given as;

K.E = ¹/₂mv²

$v_{max}^2 = \frac{2K.E}{m} \\\\v_{max}= \sqrt{\frac{2K.E}{m}} \\\\v_{max} = \sqrt{\frac{2(1.5 \ \times \ 10^{-18})}{(9.11 \ \times \ 10^{-31})}}\\\\v_{max} =1.815 \ \times \ 10^{6} \ m/s$

Therefore, the maximum speed of the ejected photon-electrons is 1.815 x 10⁶ m/s.

4 0

3 years ago