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2 years ago

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The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

y
Q(t)=t3−2t2+4t+4

Find the current when t = 1s. At what time is the current lowest?

1 answer:

Mnenie [13.5K]2 years ago

7 0

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

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The wavelength of light is 5000 angstrom. Express it in nm and m.

Ierofanga [76]

Answer:

1 angstrom = 0.1nm

5000 angstrom = 5000/1 × 0.1nm

<h3>= 500nm</h3>

$1 \: angstrom = 1 \times {10}^{ - 10} m$

5000 angstrom = 5000 × 1 × 10^-10

<h3>= 5 × 10^-7 m</h3>

Hope this helps you

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2 years ago

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$

$\omega = 2\pi (2.5)$

$\omega = 5\pi rad/s$

The angular displacement is given as the form:

$\theta (t) = \theta_0 cos(\omega t)$

In the equlibrium we have to $t=0, \theta(t) = \theta_0$ and in the given position we have to

$\theta(t) = \theta_0 cos(5\pi t)$

Derived the expression we will have the equivalent to angular velocity

$\frac{d\theta}{dt} = 2.7rad/s$

Replacing,

$\theta_0(sin(5\pi t))5\pi = 2.7$

Finally

$\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°$

Therefore the maximum angular displacement is 9.848°

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2 years ago

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

yawa3891 [41]

Answer:

Isothermal : P2 = ( P1V1 / V2 ) , work-done $pdv = nRT * In( \frac{V2}{v1} )$

Adiabatic : : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$ , work-done =

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

$pdv = nRT * In( \frac{V2}{v1} )$

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

$P1V1^y = P2V2^y$

where y = 5/3

hence : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$

Work-done

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Where $T2 = T1V1^(2/3)/V2^(2/3)$

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2 years ago

Which of the following statements is false?

agasfer [191]

A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.

Virtual images are property of concave lenses. These images appear closer but smaller than the real object.

Learn More:

For more on images formed by lenses check out;

brainly.com/question/6722295

brainly.com/question/12191285

brainly.com/question/12529812

brainly.com/question/11788630

#LearnWithBrainly

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2 years ago

Read 2 more answers

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

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2 years ago