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Ivahew [28]
2 years ago
14

Calculate the root mean square velocity of nitrogen molecules at 25°c.

Physics
1 answer:
umka2103 [35]2 years ago
5 0

Answer:

515.22 m/s

Explanation:

Formula for root mean square velocity is SQRT (3RT/M)

Where,

R = Rydberg's constant = 8.314 Kg. m²/ s². K. mol

T = Temperature in Kelvin

M = Molar mass in Kg/mol

First calculate molar mass,

As nitrogen is diatomic molecule,

Its molar mass is

M = 2(14) = 28 g/mol

Convert unit into Kg/mol

28 g/mol x 1 Kg/1000g = 2.8x10¯² Kg/mol

Convert temperature into kelvin

T = 25 + 273 = 298K

Put all values in formula

Vrms = SQRT (3RT/M)

Vrms = SQRT (3x8.314x298/2.8x10¯²)

Vrms = SQRT (265454. 14)

Vrms = 515.22 m/s

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

3 0
3 years ago
What contributions did J.J. Thomson make to atomic history?
luda_lava [24]
J.J. Thomson discovered electrons and noticed that an atom can be divided. Also, he concluded atoms are made of positive cores and negatively charged particles within it.
4 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

\displaystyle A=\frac{3}{2}B^mC^n

and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

m+n=2

-m+2n=2

Adding both equations:

3n=4

Solving:

n=4/3

m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

6 0
2 years ago
25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression
levacccp [35]

maximum speed of cheetah is

v_1 = v_{max}

speed of gazelle is given as

v_2 = v_{g}

Now the relative speed of Cheetah with respect to Gazelle

v_{12} = v_1 - v_2

v_{12} = v_{max} - v_g

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

d = v_{12}* t

so by rearranging the terms we can say

t = \frac{d}{v_{12}}

t = \frac{d}{v_{max} - v_g}

so above is the relation between all given variable

6 0
3 years ago
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
3 years ago
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