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3 years ago

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When is an atom stable? A. when it has a full outer orbit. B. when it has the same amount of elections as protons. C. when it ha

s no neutrons. D. when it has an odd number of valance electrons

2 answers:

LekaFEV [45]3 years ago

4 0

The answer is A hope this helps

balu736 [363]3 years ago

3 0

B when the electrons and protons are the same amount

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2 years ago

A lorry of mass 4000 kg is travelling at a speed of 4 m/s.

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KE = 32,000J

v = 8m/s

Explanation:

KE = .5*m*v²

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2 years ago

three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,

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Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

$C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF$

Capacitance of capacitor 2,

$C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF$

Capacitance of capacitor 3,

$C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF$

(b) The equivalent capacitance in series combination is :

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF$

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3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago