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3 years ago

How can we calculate the e.m.f of the battery?.

Physics

2 answers:

cluponka [151]3 years ago

7 0

Explanation:

The emf is equal to the work done on the charge per unit charge (ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt (1V=1J/C).

Tems11 [23]3 years ago

6 0

Answer:

The EMF or electromotive force is the energy supplied by a battery or a cell per coulomb (Q) of charge passing through it. The magnitude of emf is equal to V (potential difference) across the cell terminals when there is no current flowing through the circuit. (byjus)

Explanation:

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Prove(show) ''T=2π√(l/g)''

Nonamiya [84]

Answer:

Time period for Simple pendulum, $T=2\pi\sqrt{\frac{l}{g}$

Explanation:

The Simple Pendulum

Consider a small bob of mass $m$ is tied to extensible string of length $l$ that is fixed to rigid support. The bob is oscillating in the plane about verticle.

Let $\theta$ is the angle made by string with vertical during oscillation.

Vertical component of the force on bob, $F=-mg\sin\theta$

Negative sign shows that its opposing the motion of bob.

Taking $\theta$ as very small angle then, $\sin\theta\sim\theta$

$F=-mg\theta$

Let $x$ is the displacement made by bob from its mean position ,

then, $\theta=\frac{x}{l}$

so, $F=-mg\frac{x}{l}$ ........(1)

Since, pendulum is in hormonic motion,

as we know, $F=-kx$

where $k$ is the constant and $k=m\omega^{2}$

$F=-m\omega^2x$ .........(2)

From equation (1) and (2)

$-m\omega^2x=-mg\frac{x}{l}$

$\omega=\sqrt{\frac{g}{l}}$

Since, $\omega=\frac{2\pi}{T}$

$\frac{2\pi}{T}=\sqrt{\frac{g}{l}$

$T=2\pi\sqrt{\frac{l}{g}}$

6 0

3 years ago

a car traveling at a speed of 72km/h is uniformly returded and comes to rest after 8 seconds if the car with the occupants has a

cricket20 [7]

Answer:

the breaking force is 12

Explanation:

7 0

3 years ago

An airplane traveling at half the speed of sound emits a sound of frequency 4.68 kHz. (a) At what frequency does a stationary li

kkurt [141]

Answer:

(a) 9.36 kHz

(b) 3.12 kHz

Explanation:

(a)

V = speed of sound

$v$ = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

$f' = \frac{Vf}{V-v}$

$f' = \frac{V(4680)}{V-(0.5)V)}$

f' = 9360 Hz

f' = 9.36 kHz

(b)

V = speed of sound

$v$ = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

$f' = \frac{Vf}{V+v}$

$f' = \frac{V(4680)}{V+(0.5)V)}$

f' = 3120 Hz

f' = 3.12 kHz

6 0

3 years ago

Please help with Physics Circuits!

Zigmanuir [339]

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, $R_2, R_3, R_4$ :
$\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}$
From which we find
$R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
$R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega$
So, the correct answer is D)

2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$

And these are connected in series with a resistor of $10 \Omega$ , so the equivalent resistance of the circuit is
$R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega$

And by using Ohm's law we find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A$
So, the correct answer is C).

3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
$R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega$

And by using Ohm's law we find the current flowing in the circuit:
$I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A$

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
$V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V$
So, the correct answer is D).

4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}$
From which we find
$R_{23} = 7.55 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is:
$R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega$

The current in the circuit is given by Ohm's law
$I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A$

Now we can compare the voltage drops across the resistors. Resistor 1:
$V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V$
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
$V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V$
Resistor 4:
$V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V$

So, the greatest voltage drop is on resistor 1, so the correct answer is D).

5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
$P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W$
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

5 0

2 years ago

Read 2 more answers

Consider the reaction data. A ⟶ products T ( K ) k ( s − 1 ) 225 0.385 525 0.635 What two points should be plotted to graphicall

lutik1710 [3]

Answer:

Plot ln K vs 1/T

(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol

Explanation:

This is an example of the Arrhenius equation:

$k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = - \dfrac{E_{a}}{R}\dfrac{1}{T} + \ln A\\\\y = mx + b$

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

Data:

$\begin{array}{cccc}\textbf{k/s}\mathbf{^{-1}} &\mathbf{\ln k} & \textbf{T/K} & \mathbf{1/T(K^{-1})}\\0.285 & -0.9545 & 225 &0.004444\\0.635 & -0.4541 & 525 & 0.001905\\\end{array}$

Calculations:

(a) Rise

Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004

(b) Run

Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹

(c) Slope

Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹

(d) Activation energy

Slope = -Eₐ/R

Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol

4 0

3 years ago