Maniac Muslim Forums

## Recommended Posts

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?

##### Share on other sites

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?

##### Share on other sites

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

##### Share on other sites
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

##### Share on other sites
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

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account. ×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.