phys doubt

[halosh92]

its A

i think its short circuiting  so voltage is zero

[astarmathsandphysics]

If you look v closely they are the same circuit

[halosh92]

If you look v closely they are the same circuit

alright , in the question they say voltage is zero
how come?? is it liek each resistor in parallel to the other??

[vanibharutham]

Wow, that was a fun question

took a while:

Heres how you do it,

You need to firstly work out the Resistance of the resistor from the first circuit...
BE CAREFUL WHERE THEY HAVE PLACED THE AMMETER...ITS ON THE PARALLEL CIRCUIT... so the total current in the circuit is 2 times the ammeter reading i.e. 2A

V = I R-total
6 = 2 x R-total
R-total = 3 ohms

Now use the resistance in parallel combination:

1 / R-total = 1/2R + 1/2R

you will get:

R-total = R

therefore,

each resistor has a value of 3 ohms



in the second circuit,

V = I R
6 = I ( 3 + 3)
I = 1.0 Amps


and voltmeter must read 0 Volts because there is no emf in the path across which the voltmeter is connected

[halosh92]

Wow, that was a fun question

took a while:

Heres how you do it,

You need to firstly work out the Resistance of the resistor from the first circuit...
BE CAREFUL WHERE THEY HAVE PLACED THE AMMETER...ITS ON THE PARALLEL CIRCUIT... so the total current in the circuit is 2 times the ammeter reading i.e. 2A

V = I R-total
6 = 2 x R-total
R-total = 3 ohms

Now use the resistance in parallel combination:

1 / R-total = 1/2R + 1/2R

you will get:

R-total = R

therefore,

each resistor has a value of 3 ohms



in the second circuit,

V = I R
6 = I ( 3 + 3)
I = 1.0 Amps


and voltmeter must read 0 Volts because there is no emf in the path across which the voltmeter is connected

THANKYOU ALOOOOTTT

[Phosu]

10 Two spheres approach each other along the same straight line. Their speeds are u1 and u2
before collision, and v1 and v2 after collision, in the directions shown below.
before collision
 ->  <-
 O    O
 u1   u2

after collision
->   ->
 O    O
 v1   v2

Which equation is correct if the collision is perfectly elastic?
A u1 – u2 = v2 + v1
B u1 – u2 = v2 – v1
C u1 + u2 = v2 + v1
D u1 + u2 = v2 – v1

please explain, i ticked A, but the ans is D

[astarmathsandphysics]

For perfectly elastic collision approach speed u1+u2=separations speed v2-v1

[halosh92]

For perfectly elastic collision approach speed u1+u2=separations speed v2-v1

is this ALWAYS the formula ?
and could you please explain when do we use these formulas exactly?
very urgent plz and wat s speed of approach and separation anywway
thxx aloot
VERY URGENT PLZ

[astarmathsandphysics]

Use the formula if you know a collision is perfectly elastic.
Speed of approach is the closing speed - add if they are moving towards each other, subtract if they are moving in the same direction.
Separation speed is the speed at which they are parting. Add if they are moving away from each other, subtract if they are moving in the same direction

[Phosu]

The interference patterns from a diffraction grating and a double slit are compared.
Using the diffraction grating, yellow light of the first order is seen at 30° to the normal to the
grating.

The same light produces interference fringes on a screen 1.0 m from the double slit. The slit
separation is 500 times greater than the line spacing of the grating.

What is the fringe separation on the screen?

A 2.5 × 10^–7m
B 1.0 × 10^–5m
C 1.0 × 10^–3m
D 1.0 × 10^–1m

plz show how to solve,
i know the two formulas, but i dont know how to use them

[halosh92]

The interference patterns from a diffraction grating and a double slit are compared.
Using the diffraction grating, yellow light of the first order is seen at 30° to the normal to the
grating.

The same light produces interference fringes on a screen 1.0 m from the double slit. The slit
separation is 500 times greater than the line spacing of the grating.

What is the fringe separation on the screen?

A 2.5 × 10^–7m
B 1.0 × 10^–5m
C 1.0 × 10^–3m
D 1.0 × 10^–1m

plz show how to solve,
i know the two formulas, but i dont know how to use them

is it C?

[astarmathsandphysics]

When I get home

[Phosu]

is it C?

yah, how did you find it out?

[halosh92]

yah, how did you find it out?

ok so first we have these 2 formulas:

d sin thita= n (lambda)              equ 1
(lambda) = ax/D                       equ 2

the information we got for equation one is :
thita=30
n=1 (first order as mentioned)
d= unknown
wavelength of yellow light approximately = 500nm


the information we got for equ 2 is:
D=1 m
a = 500 d ( as they say its 500 times)

therefore substitute values in equ 1 :
d sin thita= n lambda
rearrange
d= (1 * (500 *10^-9) )/ sin 30                    >>>>> (mind u we multiply 500nm * 10^-9 to change it to "m" )
d= 1 * 10^-6 m


now substitute in equ 2 :

lambda= ax/D
rearrange
lambda/aD =x

now substitue values:
( 500 * 10^-9)/(500 * (1*10^-6)  *1)

answer:
1 *10^-3   which is C

[Phosu]

Thanks so much for the explanation

a few doubts, do we need to know the wavelength of each light?

i know n stands for the order, but what does "d" stand for?

and you got the last rearrangement rong,
if (lambda)= ax/D
then (lambda)D/a=x
in this case since D is equal to 1, it wouldn't affect the answer. but in other cases be careful.