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3 years ago

10

A conductor carries a current that is decreasing exponentially with time. The current is modeled as ????=????0????−????/???? , w

here ????0=3.00A is the current at time ????=0.00s and ????=0.50s is the time constant. How much charge flows through the conductor between ????=0.00s and ????=3???? ?

1 answer:

Vlad1618 [11]3 years ago

3 0

Answer:

Q=1.42 C

Explanation:

Given that

$I=I_oe^{\dfrac{-t}{\tau}}$

When t= 0 ,Io = 3 A

τ = Time constant = 0.5 s

$I=3e^{-\dfrac{t}{\tau}}$

We know that

$I=\dfrac{dQ}{dt}$

Q=Charge ,I =Current

Q = ∫I.dt

Given that

t= 0 to t= 3τ= 1.5 s

The charge Q

$Q=\int_{0}^{1.5}3e^{-\dfrac{t}{0.5}}dt$

Q=1.42 C

Therefore charge flow conductor will be 1.42 C.

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Which best describes the motion of air particles when a transverse wave passes through them?

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C.
The particles move perpendicular to the direction of the wave.

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3 years ago

The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b

bogdanovich [222]

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

8 0

3 years ago

A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total

Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

$v_2 = v - 1$

now we know that total time of the motion is 19 hr

so we will have

$t_1 = \frac{90}{v}$

$t_2 = \frac{90}{v - 1}$

so we will have

$t_1 + t_2 = 19 hr$

$\frac{90}{v} + \frac{90}{v-1} = 19$

$90(2v - 1) = 19(v^2 - v)$

$19 v^2 - 199 v + 90 = 0$

by solving above equation we have

$v = 10 mph$

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0

3 years ago

S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(

Ket [755]

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.

V = kq/r

Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

To learn more about electric potential refer to:

brainly.com/question/12645463

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3 0

2 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

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3 years ago