Tavaline to be ali dvedenen Mona and Sauna How much will eacher
to be divided sually into thwart murh
Answers
Corona Go Is What You Want To Say Right ?? Ramdas Athwale Said That Corona Go - Corona Go - Go Corona
Answer:
Answer:
Correct Question -
The circumference of two circle are in the ratio 2 : 3. Find the ratio of their areas.
Given -
Ratio of their circumference = 2:3
To find -
Ratio of their areas.
Formula used -
Circumference of circle
Area of circle.
Solution -
In the question, we are provided, with the ratios of the circumference of 2 circles, and we need to find the ratio of area of those circle, for that first we will use the formula of circumference of a circle, then we will use the formula of area of circles. We will be writing 1 equation in it too.
So -
Let the circumference of 2 circles be c1 and c2
According to question -
c1 : c2
Circumference of circle = 2πr
where -
π = \tt\dfrac{22}{7}722
r = radius
On substituting the values -
c1 : c2 = 2 : 3
2πr1 : 2πr2 = 2 : 3
\tt\dfrac{2\pi\:r\:1}{2\pi\:r\:2}2πr22πr1 = \tt\dfrac{2}{3}32
\tt\dfrac{r1}{r2}r2r1 = \tt\dfrac{2}{3}32 \longrightarrow⟶ [Equation 1]
Now -
Let the areas of both the circles be A1 and A2
Area of circle = πr²
So -
Area of both circles = πr1² : πr2²
On substituting the values -
A1 : A2 = πr1² : πr2²
\tt\dfrac{A1}{A2}A2A1 = \tt\dfrac{(\pi\:r1)}{(\pi\:r2)}^{2}(πr2)(πr1)2
\tt\dfrac{A1}{A2}A2A1 = \tt\dfrac{(r1)}{(r2)}^{2}(r2)(r1)2
\tt\dfrac{A1}{A2}A2A1 = \tt\dfrac{(2)}{(3)}^{2}(3)(2)2 [From equation 1]
So -
\tt\dfrac{A1}{A2}A2A1 = \tt\dfrac{4}{9}94
\therefore∴ The ratio of their areas is 4 : 9
______________________________________________________
you would abhor to do me wrong as much as i to spoil your song" could the speaker actually spoil the song? why/why not?... Follow me......
Here is ur ans....
Given
AD=BC and AD∥BC
To prove:
AB=DC
Proof:
In △ABC and △CDA
BC=DA (given)
∠BCA=∠DAC (alternate angles)
AC=AC (common)
∴△ABC≅△CDA [by SAS congruence property]
Hence AB=CDThe Nightingale and the Glow-Worm
by
William Cowper

Next
A nightingale, that all day long
Had cheered the village with his song,
Nor yet at eve his note suspended,
Nor yet when eventide was ended,
Began to feel, as well he might,
The keen demands of appetite;
When, looking eagerly around,
He spied far off, upon the ground,
A something shining in the dark,
And knew the glow-worm by his spark;
So, stooping down from hawthorn top,
He thought to put him in his crop;
The worm, aware of his intent,
Harangued him thus right eloquent:
"Did you admire my lamp," quoth he,
"As much as I your minstrelsy,
You would abhor to do me wrong,
As much as I to spoil your song,
For 'twas the self-same power divine
Taught you to sing, and me to shine,
That you with music, I with light,
Might beautify and cheer the night."
The songster heard his short oration,
And warbling out his approbation,
Released him, as my story tells,
And found a supper somewhere else.
2.5%..
hope it helps u
mark me as brainliestAnswer:
\bullet\;\;\boxed{\sf E=\dfrac{r}{2}\;.\;\bigg(\dfrac{dB}{dt}\bigg)}∙E=2r.(dtdB)
Explanation:
\rule{300}{1.5}
As given the Magnetic field is increasing at a rate of dB/dt, therefore the magnetic field is not constant. Ans we need to find electric field inside the circular region as r < R. So, from the relation we know that,
\begin{gathered}\\\end{gathered}
\longrightarrow\sf V=\displaystyle\oint\sf E\;.\;ds⟶V=∮E.ds
Here,
V Denotes Voltage/EMF.
E Denotes electric field.
ds denotes small elemental length/distance.
Solving further,
\begin{gathered}\longrightarrow\sf \dfrac{d \phi}{dt}=\displaystyle\oint\sf E\;.\;ds\ \ \ \ \ \because\Bigg[\dfrac{d\phi}{dt}=V\Bigg]\\\\\\\\\longrightarrow\sf \dfrac{d \phi}{dt}=E\displaystyle\oint\sf ds\\\\\\\\\longrightarrow\sf \dfrac{d\big(B\;.\;A\big)}{dt}=E\displaystyle\oint\sf ds\\\\\\\\\longrightarrow\sf E\displaystyle\oint\sf ds=A\;.\;\dfrac{dB}{dt}\\\\\\\\\longrightarrow\sf E\displaystyle\oint\sf ds=\pi r^{2}\;.\;\dfrac{dB}{dt}\end{gathered}⟶dtdϕ=∮E.ds ∵[dtdϕ=V]⟶dtdϕ=E∮ds⟶dtd(B.A)=E∮ds⟶E∮ds=A.dtdB⟶E∮ds=πr2.dtdB
∫ ds = 2πr as it will cover whole length of the circle i.e. circumference.
\begin{gathered}\longrightarrow\sf E\times 2\pi r=\pi r^{2}\;.\;\dfrac{dB}{dt}\\\\\\\\\longrightarrow\sf E=\dfrac{\pi r^{2}}{2\pi r}\;.\;\Bigg(\dfrac{dB}{dt}\Bigg)\\\\\\\\\longrightarrow\sf E=\dfrac{r}{2}\;.\;\Bigg(\dfrac{dB}{dt}\Bigg)\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf E=\dfrac{r}{2}\;.\;\Bigg(\dfrac{dB}{dt}\Bigg)}}}}\end{gathered}⟶E×2πr=πr2.dtdB⟶E=2πrπr2.(dtdB)⟶E=2r.(dtdB)⟶E=2r.(dtdB)
\begin{gathered}\\\end{gathered}
Hence, we got the value of electric field.
\rule{300}{1.5}