please help help me please
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Hi there,
1. given: h1(object size) = 2 cm, u (object distance) = -20cm, f (focal length) = -12 cm
By mirror formula:
1/v + 1/u = 1/f
1/v = 1/f - 1/u
= 1/(-12) - 1/(-20)
=( -5 + 3)/60. (common denominator)
= 8/60
= 7.5 cm
therefore image distance (v) = 7.5 cm
magnification formula:
M = h2/h1 = (-v)/u ( h2 is image height and h1 is object height)
h2 = - (v x h1)/u
h2 = - [(7.5)(2)] / (-20)
h2 = 0.75 cm
therefore the height of the image is 0.75cm and it is an erect and diminished image.
hope it helps you out mate!
2. Formula :
Work (W) = Potential difference (V) x Charge (Q)
W = 8 x 12
W = 96 joules.
therefore, work done is 96J
3. Formula:
Potential difference (V) = Current (I) x Resistance (R)
24 = I x 6
I = 4 Amperes.
therefore, the current drawn by the bulb is 4A.
sorry mate but I don't know the 4th question.
still I hope it helps you out mate
1. given: h1(object size) = 2 cm, u (object distance) = -20cm, f (focal length) = -12 cm
By mirror formula:
1/v + 1/u = 1/f
1/v = 1/f - 1/u
= 1/(-12) - 1/(-20)
=( -5 + 3)/60. (common denominator)
= 8/60
= 7.5 cm
therefore image distance (v) = 7.5 cm
magnification formula:
M = h2/h1 = (-v)/u ( h2 is image height and h1 is object height)
h2 = - (v x h1)/u
h2 = - [(7.5)(2)] / (-20)
h2 = 0.75 cm
therefore the height of the image is 0.75cm and it is an erect and diminished image.
hope it helps you out mate!
2. Formula :
Work (W) = Potential difference (V) x Charge (Q)
W = 8 x 12
W = 96 joules.
therefore, work done is 96J
3. Formula:
Potential difference (V) = Current (I) x Resistance (R)
24 = I x 6
I = 4 Amperes.
therefore, the current drawn by the bulb is 4A.
sorry mate but I don't know the 4th question.
still I hope it helps you out mate
AtharvaMalji:
you know what, in our textbook we got focal length of concave mirror negative,
LOL
i am correcting it.
but in my text book focal length and radius of concave mirror is taken as positive and convex as positive.
dont mind from mistakes we learn
my concave mirror is facing towards left(negative) side of principal axis in my textbook.
the reflecting side should face on right
that's the catch. we have new Cartesian sign conventions which states this.
check it now I tried to do something around.
just correct me where I am wrong.
Answered by
0
for question 2
given,
charge(q)=12C
pottential difference(V)=8 volt
work(w)=?
now,
from the defination of pottential difference,
V=W/Q
or,W=V×Q
or,W=12×8
or,W=96 j
therefore, the required amount of workdone is 96j
for question no 4
magnification(m)=-3 (erect image, it is virtual)
if image distance,u=x
then,magnification(m)=v/u
or,-3=v/u
so, image distance,u=-3x
for concave mirror,R=36cm
so,f=R/2
=36/2
=18cm
now using mirror formula,
i.e 1/f=1/u+1/v
or, 1/18=1/x+(-1/3x)
or, 1/18=3-1/3x
or, 1/18=2/3x
therefore, x=12cm
so, the object is situated 12cm infront of the mirror
i think this will help you question no 4 previous friend have already solved and 1 i am confused.....
given,
charge(q)=12C
pottential difference(V)=8 volt
work(w)=?
now,
from the defination of pottential difference,
V=W/Q
or,W=V×Q
or,W=12×8
or,W=96 j
therefore, the required amount of workdone is 96j
for question no 4
magnification(m)=-3 (erect image, it is virtual)
if image distance,u=x
then,magnification(m)=v/u
or,-3=v/u
so, image distance,u=-3x
for concave mirror,R=36cm
so,f=R/2
=36/2
=18cm
now using mirror formula,
i.e 1/f=1/u+1/v
or, 1/18=1/x+(-1/3x)
or, 1/18=3-1/3x
or, 1/18=2/3x
therefore, x=12cm
so, the object is situated 12cm infront of the mirror
i think this will help you question no 4 previous friend have already solved and 1 i am confused.....
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