Answer:
Explanation:
- For diagram refer the attachment.
It is given that five cells of 2V are connected in series, so total voltage of the battery:
![\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%5C%3A%20%5C%3A%20%20%5Csf%20V%20%3D%202%20%5Ctimes%205%20%3D%20%3Cstrong%3E10%20V%3C%2Fstrong%3E%20)
Three resistor of 5
, 10
, 15
are connected in Series, so the net resistance:
![\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%5C%3A%20%5C%3A%20%5Csf%20R_%7Bn%7D%20%3D%20R_%7B1%7D%20%2B%20R_%7B2%7D%20%2B%20R_%7B3%7D)
![\dashrightarrow \: \: \sf R = 5 + 10 + 15](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%5C%3A%20%5C%3A%20%20%5Csf%20R%20%3D%205%20%2B%2010%20%2B%2015%20)
![{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}](https://tex.z-dn.net/?f=%7B%20%5Cpink%7B%5Cdashrightarrow%20%5Csf%20%5C%3A%20%5C%3A%20%7B%20%5Cunderbrace%7BR%20%3D%2030%20%5C%3A%20%20%5COmega%7D%7D%7D%7D)
According to ohm's law:
![\dashrightarrow \sf\: \: V = IR](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%20%5Csf%5C%3A%20%5C%3A%20V%20%3D%20IR%20)
![\dashrightarrow \sf \: \: I = \dfrac{V}{R}](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%20%5Csf%20%5C%3A%20%5C%3A%20I%20%3D%20%5Cdfrac%7BV%7D%7BR%7D)
On substituting resultant voltage (V) as 10 V and resultant resistant, as 30
we get:
![\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%5Csf%20%5C%3A%20%5C%3A%20I%20%3D%20%5Cdfrac%7B10V%7D%7B30%5COmega%7D)
![{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}](https://tex.z-dn.net/?f=%7B%20%5Cpink%7B%5Cdashrightarrow%20%5Csf%20%5C%3A%20%5C%3A%20%7B%20%5Cunderbrace%7BI%20%3D%200.33%20A%7D%7D%7D%7D)
The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>
Answer:
1.58 Hz
Explanation:
The frequency of the simple pendulum is given by
f = 1/T
= 1/2π√g/l
In this problem, I = 10.0 cm = 0.1 m
f = 1/2π√9.8/0.1
= 1.58 Hz
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Answer:
<em>the ball travels a distance of 8.84 m</em>
Explanation:
Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.
R = (U²sin2∅)/g.............................. Equation 1
Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.
<em>Given: U = 10 m/s, ∅ = 60°</em>
<em>Constant: g = 9.8 m/s²</em>
Substituting these values into equation 1
R = [10²×sin(2×60)]/9.8
R = (100sin120)/9.8
R = 100×0.8660/9.8
R = 86.60/9.8
R = 8.84 m
<em>Therefore the ball travels a distance of 8.84 m</em>
Answer:
(a) ![emf_L=-LI_{max}\omega cos(\omega t)](https://tex.z-dn.net/?f=emf_L%3D-LI_%7Bmax%7D%5Comega%20cos%28%5Comega%20t%29)
(b) neither increasing or decreasing
(c) opposite to the flow of charge carriers
Explanation:
The current through an inductor of inductance L is given by:
(1)
(a) The induced emf is given by the following formula
(2)
You derivative the expression (1) in the expression (2):
![emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)](https://tex.z-dn.net/?f=emf_L%3D-L%5Cfrac%7Bd%7D%7Bdt%7D%28I_%7Bmax%7Dsin%28%5Comega%20t%29%29%5C%5C%5C%5Cemf_L%3D-LI_%7Bmax%7D%5Comega%20cos%28%5Comega%20t%29)
(b) At t=0 the current is zero
(c) At t = 0 the emf is:
![emf_L=-\omega LI_{max}](https://tex.z-dn.net/?f=emf_L%3D-%5Comega%20LI_%7Bmax%7D)
w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.
(d) read the text carefully