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3 years ago

7

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele

ctrons from an initial point where the electric potential is 6.00 V to a point where the potential is -5.00 V

1 answer:

igomit [66]3 years ago

8 0

Answer:

$1.06\times 10^6 J$

Explanation:

We are given that

$V_1=6 V$

$V_2=-5 V$

Charge on 1 electron, 1 e= $-1.6\times 10^{-19} C$

Avogadro's number, $N_A=6.02\times 10^{23}$

$q=ne$

$q=6.02\times 10^{23}\times 1.6\times 10^{-19}=-9.632\times 10^4 C$

$W=q\Delta V$

$W=q(V_2-V_1)$

$W=-9.632\times 10^4\times (-5-6)=1.06\times 10^6 J$

Hence, the work done= $1.06\times 10^6 J$

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A ball is spun around in circular motion such that it completes 50 rotations in 25 s. What is the frequency of its rotation? 2.

Elis [28]

Answer:

1. $f = 2 Hz$

2. $f = 0.011 Hz$

3. $f = 0.067 Hz$

4. $t = 4 s$

Explanation:

1. The frequency of rotation is given by:

$f = \frac{\omega}{2\pi}$

Where:

ω: is the angular speed = 50 rotations (revolutions) in 25 s.

We need to convert the units of ω.

$\omega = \frac{50 rev}{25 s}*\frac{2\pi rad}{1 rev} = 4\pi rad/s$

Now, the frequency is:

$f = \frac{4\pi rad/s}{2\pi} = 2 Hz$

2. The frequency is:

We know:

5 laps = 5 revolutions

t: time = 450 s

$f = \frac{\omega}{2\pi} = \frac{\frac{5 rev}{450 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.011 Hz$

3. The frequency of the pendulum is:

$f = \frac{\omega}{2\pi} = \frac{\frac{1 rev}{15 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.067 Hz$

4. We have:

θ: number of revolutions = 48 rev

f = 12 Hz

t =?

The time can be calculated as follows:

$f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t}$

$t = \frac{\theta}{2\pi f} = \frac{48 rev*\frac{2\pi rad}{1 rev}}{2\pi*12 Hz} = 4 s$

I hope it helps you!

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3 years ago

When you lift a book from the ground to your desk, what kind of work do you do negative or positive by lifting the book what do

skelet666 [1.2K]

It would be kinetic energy. Let's say the book is weighs 10 Newtons you need to use a force of 10 Newtons to lift the book. In other words it's positive. As you move the book you're giving it energy. Namely potential energy which will turn to kinetic energy if you let it go. So you're changing it's position and energy.

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4 years ago

Noah is loading the ark and the last animal on board is a stubborn 1500-kg elephant who refuses to budge. Noah and his family pu

Oxana [17]

The coefficient of sliding friction is 0.514

Explanation:

We start by writing the equations of motion of the elephant along the two directions, parallel and perpendicular, to the incline.

Along the parallel direction we have:

$F- mg sin \theta - \mu_k R = ma$ (1)

where :

F = 10,000 N is the force applied by Noah

$mg sin \theta$ is the component of the weight parallel to the incline, where:

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=10^{\circ}$ is the angle of incline

$\mu_k R$ is the force of friction, where:

$\mu_k$ is the coefficient of friction

R is the normal reaction

and a is the acceleration

Perpendicular direction:

$R-mg cos \theta =0$ (2)

where $mg cos \theta$ is the component of the weight perpendicular to the incline

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$F-mg sin \theta - \mu_k mg cos \theta = ma$

We know that the elephant moves at constant speed, so the acceleration is zero:

a = 0

So the equation becomes

$F-mg sin \theta - \mu_k mg cos \theta=0$

And we can re-arrange it to find the coefficient of friction:

$F-mg sin \theta - \mu_k mg cos \theta=0\\\mu_k = \frac{F-m g sin \theta}{mg cos \theta}=\frac{10000-(1500)(9.8)(sin 10)}{(1500)(9.8)(cos 10)}=0.514$

Learn more about friction and inclined planes:

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3 years ago

Stewart (70 Kg) is attracted to Ms. Little (60 Kg) who sits 2 m away. What is the gravitational attraction between them? G=6.67×

ikadub [295]

Happy Holidays!

We can use the following equation to solve for the gravitational force:

$\large\boxed{F_g = G\frac{m_1m_2}{r^2}}$

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G = Gravitational constant

m1,m2 = masses of the objects (kg)

r = distance between the objects (m)

Plug in the given values into the equation:

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Yurem is pulling a wagon across the playground with a force of 10 N. He asks Elianna to help. She agrees and pushes the back of

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Answer:

22 N applied force

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3 years ago

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