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4 years ago

7

Why do heavier objects fall at the same speed as lighter ones?

1 answer:

GaryK [48]4 years ago

5 0

Speed of any freely falling object is always same. Provided, both are left to fall from the same height. If you perform this experiment in a perfect vacuum or near vacuum laboratory, both of them will reach ground with same velocity this is because there is no resistance to their motion. This is always true no matter where you go and perform this experiment.
It can be easily proved from conservation of mechanical energy. Why conserving energy? because there are no forces acting on the freely falling objects other than conservative force(mg).

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Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV

forsale [732]

Answer:

(a) $\lambda=1.227\ A$

(b) $\lambda=0.388\ A$

(c) $\lambda=0.038\ A$

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where

m and e are the mass of and charge on an electron

On solving,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

V = 100 V

$\lambda=\dfrac{12.27}{\sqrt{100} }\ A$

$\lambda=1.227\ A$

(b) V = 1 kV = 1000 V

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{1000} }\ A$

$\lambda=0.388\ A$

(c) If $V=100\ kV=10^5\ V$

$\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A$

$\lambda=0.038\ A$

Hence, this is the required solution.

7 0

3 years ago

18°C = _____ <br>-255 K<br> 0 K <br>18 K <br>291 K

Nadya [2.5K]

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3 years ago

Read 2 more answers

Two particles of equal mass m are at the vertices of the base of an equilateral triangle. The triangle’s center of mass is midwa

Helga [31]

Answer:Twice of given mass

Explanation:

Given

Two Particles of Equal mass placed at the base of an equilateral Triangle

let mass of two equal masses be m and third mass be m'

Taking one of the masses at origin

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Co-ordinates of other equal mass is (a,0)

if a is the length of triangle

co-ordinates of final mass $(\frac{a}{2},\frac{\sqrt{3}a}{2})$

Given its center of mass is at midway between base and third vertex therefore

$x_{cm},y_{cm}=\frac{a}{2},\frac{\sqrt{3}a}{4}$

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$\frac{\sqrt{3}a}{4}=\frac{m\cdot 0+m\cdot 0+m'\cdot \frac{\sqrt{3}a}{2}}{m+m+m'}$

$2m+m'=4\times (\frac{m'}{2})$

$2m+m'=2m'$

$m'=2m$

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3 years ago

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KonstantinChe [14]

Answer:

1. OK the

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And

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(4)

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CaHeK987 [17]

Answer: 0.0 m/s

Explanation

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3 years ago