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3 years ago

PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS

Physics

1 answer:

butalik [34]3 years ago

8 0

Answer:

g=GM/R^2

Universal Gravutation Constant:

f=GM×m/R^2

Force can be also expressed as

f=m×g

so,

mg=GMxm/R^2

The m gets cancelled so

g=GM/R^2

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She could look for an online job or a job she can work at from home.

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2 years ago

Read 2 more answers

A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou

Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

$\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m$

$\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s$

State 2 : constant speed

$\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m$

State 3 : deceleration

$\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)$

$\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m$

Total distance : state 1+ state 2+state 3

$\tt 200 + 36000 + 300=36500~m$

the average velocity = total distance : total time

$\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s$

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2 years ago

How do two sublevels of the same principal energy level differ from each other

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Each energy sublevel corresponds to an orbital of a different shape.

Explanation:

Two sublevels of the same principal energy level differs from each other if the sublevels corrresponds to an orbital of a different shape.

The principal quantum number of an atom represents the main energy level in which the orbital is located or the distance of an orbital from the nucleus. It takes values of n = 1,2,3,4 et.c
The secondary quantum number gives the shape of the orbitals in subshells accommodating electrons.
The number of possible shapes is limited by the principal quantum numbers.

Take for example, Carbon:

1s² 2s² 2p²

The second energy level is 2 but with two different sublevels of s and p. They have different shapes. S is spherical and P is dumb-bell shaped .

Learn more:

Quantum number brainly.com/question/9288609

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3 years ago

which force field can accelerate an electron, but never change its speed? a) electric field b) magnetic field c) both of these d

Drupady [299]

Option a; Electric field can accelerate an electron, but never change its speed

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1 year ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS​

PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS