Question

In what endeavour does
Mrs.
Fitzgerald help Mrs. Pearson?
​

Answer 1

Answer:

p>Time=1hour</p><p></p><p>ToFind</p><p></p><p>Charge</p><p></p><p>Solution</p><p></p><p>Weknowthat</p><p></p><p>⟶Q=It⟶Q=It</p><p></p><p>Here</p><p></p><p>Q=Charge</p><p></p><p>I=Current</p><p></p><p>t=Time</p><p></p><p>Units</p><p></p><p>Q=Coulomb(C)</p><p></p><p>I=Ampere(A)</p><p></p><p>t=seconds(s)</p><p></p><p>Accordingthequestion</p><p></p><p>Weareaskedtofindcharge</p><p></p><p>Therefore</p><p></p><p>Wemustfind"Q"</p><p></p><p>Giventhat</p><p></p><p>Current=0.4A</p><p></p><p>Time=1hour</p><p></p><p>Hence</p><p></p><p>I=0.4A</p><p></p><p>t=1hr=3600s</p><p></p><p>Substitutingthevalues</p><p></p><p>Weget</p><p></p><p>⟶Q=0.4×3600 C⟶Q=0.4×3600 C</p><p></p><p>Therefore</p><p></p><p>⟶Q=1440 C⟶Q=1440 C</p><p></p><p>Hence</p><p></p><p>Charge=1440 CCharge=1440 C<Given:−</p><p></p><p>Focallength=12cm</p><p></p><p>Distance=−12cm</p><p></p><p>Height=7cm</p><p></p><p>ToFind:−</p><p></p><p>Positionandnatureandheight</p><p></p><p>Solution :−</p><p></p><p>Asweknowthat</p><p></p><p>

f

1

=

v

1

+

u

1

f1=v1+u1</p><p></p><p>

12

1

=

v

1

+

−12

1

121=v1+−121</p><p></p><p>

12

1

=

v

1

+

12

1

121=v1+121</p><p></p><p>

v

1

=

12

1

−

12

1

v1=121−121</p><p></p><p>

v

1

=

12

1−1

=

12

0

v1=121−1=120</p><p></p><p>v=12v=12</p><p></p><p>Hence :−</p><p></p><p>Theheightofimageis12cm</p><p></p><p>Now,</p><p></p><p>Let

′

sfindmagnification</p><p></p><p>

hi

ho

=

u

−v

hiho=u−v</p><p></p><p>

hi

7

=

−12

−12

hi7=−12−12</p><p></p><p>

hi

7

=1hi7=1</p><p></p><p>hi=7cmhi=7cm</p><p></p><p>Hence :−</p><p></p><p>Heightofimageis7cmanditisvirtualanderect.</p><p></p><p>$\huge\red{\underline\orange{\underline\purple{......}}}$