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Salaam,

 

So i got a question, its more into astronomy and astrophysics but maybe those studying physics can help.

Theoretically if the universe was expanding at a slower rate(which it is not), then would the distance between the Earth and other galaxies become less or more and hence will the supernoves(type 1A) become brighter or dimmer?

 

Or to phrase it better, would the supernovas(type 1A) become more further away and dimmer if the rate of expansion of the universe was fixed?

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slower expansion = less distance

faster expansion = more

 

Hubble law: v = H d

 

 

further the illuminating object is dimmer it gets. (true for nova)

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Yo

 

In a tv, e- strike the screen after being acceerated from rest through a potential differenc of 25,000 V. ignoring the effects of special relativity at high speeds, which really should be incl. here, find the electron speed just before the electron strikes the screen.

 

so i got...

 

deltaV = 25000 V

v0 = 0 m/s

vf = UNKNOWN

Felectric is the only force on the e- so....

Fnet = Felectric = ma

m = 9.11 x 10^-31 kg

Felectric = qE

but E is not given

however, deltaV = -Ed

but d is not given

 

qE = ma

I have q, not E, have m, not a...

& I have V...

now what?

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pot. energy = Vq (V=20000, q is electron charge)

 

1/2 m v^2 = V q (m=electron mass, v=speed)

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Yo

 

The work done by an external force to move a -2.0 uC charge from pt A to pt B is 8 x 10^-4 J. if the charge was started from rest and had 1 x 10^-4 J of KE when it reached pt B, what must be the pot diff btwn A and B?

 

Va

Vb

deltaV = Vb - Va

 

q = -2.0 x 10^-6 C

W = 8 x 10^-4 J

vo = 0

 

Wexternal force = Fexternal force (d) = qEd = qdeltaV

8 x 10^-4 J = -2.0 x 10^6 C (deltaV)

deltaV = 400 V

 

400 V = Vb - Va

PEa = qVa

PEa = KEb = 1.0 x 10^-4 J = 2.0 x 10^-6 C (Va)

Va = 50 V

 

400 V = Vb - 50V

Vb = 350 V

 

So is the "potential diff btwn A and B":

* Va

* Vb

or

*deltaV

 

cuz it gives the answer as -350 V

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OH YEAH

 

PE = qV

KE = -qV = 1/2mv^2

 

negative right?

 

 

not negative, because V is also negative.

 

i.e., q = -6.1blablabla

V = 0 - 20000 (V here is potential difference)

= -20000

 

so q*V >0

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Wexternal force = Fexternal force (d) = qEd = qdeltaV

8 x 10^-4 J = -2.0 x 10^6 C (deltaV)

deltaV = -400 V (forgot minus here)

 

400 V = Vb - Va

PEa = qVa

PEa = KEb = 1.0 x 10^-4 J = 2.0 x 10^-6 C (Va)

Va = 50 V

 

400 V = Vb - 50V

Vb = 350 V

 

So is the "potential diff btwn A and B":

* Va

* Vb

or

*deltaV

 

cuz it gives the answer as -350 V

 

 

pot diff is deltaV

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Yo

 

The work done by an external force to move a -2.0 uC charge from pt A to pt B is 8 x 10^-4 J. if the charge was started from rest and had 1 x 10^-4 J of KE when it reached pt B, what must be the pot diff btwn A and B?

 

Va

Vb

deltaV = Vb - Va

 

q = -2.0 x 10^-6 C

W = 8 x 10^-4 J

vo = 0

 

Wexternal force = Fexternal force (d) = qEd = qdeltaV

8 x 10^-4 J = -2.0 x 10^6 C (deltaV)

deltaV = 400 V

 

400 V = Vb - Va

PEa = qVa

PEa = KEb = 1.0 x 10^-4 J = 2.0 x 10^-6 C (Va)

Va = 50 V

 

400 V = Vb - 50V

Vb = 350 V

dang,... what kind of a eqn solving is this

So is the "potential diff btwn A and B":

* Va

* Vb

or

*deltaV

 

cuz it gives the answer as -350 V

 

 

see red above

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i know, its just cuz i missed the minus sign

 

deltaV = -400 V

deltaV = Vb - Va

-400V = Vb - 50V

Vb = -350V

 

so i still dont get why it should be Vb and not deltaV...

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