Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one?
What is the charge, in μC, on the negative one?

Answers

Answer 1

Answer:

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

$U_(e) = (k\cdot q_(A)\cdot q_(B))/(r)$ (1)

Where:

$U_(e)$ - Electric potential energy, measured in joules.

$q_(A)$ , $q_(B)$ - Electric charges, measured in coulombs.

$r$ - Distance between charges, measured in meters.

$k$ - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that $U_(e) = -24\,J$ , $q_(A) = 45* 10^(-6)\,C+ q_(B)$ , $k = 9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2))$ and $r = 0.29\,m$ , then the electric charge is:

$-24\,J = -(\left(9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) \right)\cdot (45* 10^(-6)\,C+q_(B))\cdot q_(B))/(0.29\,m)$

$-6.96 = -405000\cdot q_(B)-9* 10^(9)\cdot q_(B)^(2)$

$9* 10^(9)\cdot q_(B)^(2)+405000\cdot q_(B) -6.96 = 0$ (2)

Roots of the polynomial are found by Quadratic Formula:

$q_(B,1) = 1.327* 10^(-5)\,C$ , $q_(B,2) \approx -5.827* 10^(-5)\,C$

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

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An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Answers

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]

Answers

The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t

Where,

θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration = 4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

$\rm \omega_f^2 - \omega_i^2 = 2as \n\n \rm \omega_f^2 - 0 = 2* 4 * 62.83 \n\n \rm \omega_f= 22.41 \ rad/sec$

Hence the angular velocity will be 22.41 rad/s.

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Answer:

$w_f$ = 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

$w_f^2-w_i^2=2aS$

where $w_f$ is the final angular velocity, $w_i$ the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

$w_f^2-(0)^2=2(4)(62.83)$

Finally, solving for $w_f$ :

$w_f$ = 22.41rad/s

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=(mv^2)/(2)+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

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The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.How many capillaries are there?

Answers

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

$A_1v_1=nA_2v_2$ (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

$A=\pi r^2\n\nA_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\n\nA_2=\pi r_2^2=\pi (5*10^(-4)cm)^2=7.85*10^(-7)cm^2$

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

$n=(A_1v_1)/(A_2v_2)=((3.1415cm^2)(40cm/s))/((7.85*10^(-7)cm^2)(0.10cm/s))=1.6*10^9$

Then, the number of capillaries is 1.6*10^9

Talia is on a road trip with some friends. In the first 2 hours, they travel 100 miles. Then they hit traffic and go only 30 miles in the next hour. The last hour of their trip, they drive 75 miles.Calculate the average speed of Talia’s car during the trip. Give your answer to the nearest whole number.

Answers

Answer:

51 mph

Explanation:

Since Speed, V = Distance/Time
Average speed = Total Distance/Total Time

From the given data, Total Distance = 100 + 30 + 75 miles
and Total Time = 2 + 1 + 1 hours

Average Speed = 205/4
Average Speed = 51.25 mph ( or 51mph to the nearest whole number)

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