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Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.

Answers

Answer 1

Answer:

Answer:

-9Q

Explanation:

Electric field at origin is:

$E=(2keQ)/(a^2)$

Electric field due to first charge at origin would be:

$E_1=(keQ)/(a^2)$

Electric field due to second charge would be:

$E_2=E-E_1\nE_2=(2keQ)/(a^2)-(keQ)/(a^2) = (keQ)/(a^2)$

If the second charge is Q', then $E_2$ should be:

$E_2=(keQ')/((3a)^2)=(keQ')/(9a^2)$

compare the above two values to find the possible values of Q':

$(|Q'|)/(9)=Q\n |Q'|=9Q$

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Answer 2

Answer:

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.

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The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .

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
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Average Speed = 205/4
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A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.39 nm . Find the electric force on the proton.Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answers

Answer:

The electric force on the proton is 8.2x10^-10 N

Explanation:

We use the formula to calculate the distance between two points, as follows:

r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:

r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m

We will use the following expression to calculate the electrostatic force:

F = (q1*q2)/(4*pi*eo*r^2)

Here we have:

q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2

Replacing values:

F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

Answers

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

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Velocity acquired during this time

$\Rightarrow v_x=4.3+4.6* 1.44\n\Rightarrow v_x=4.3+6.624\n\Rightarrow v_x=10.92\ s$

Consider vertical motion

$\Rightarrow v_y=0+7(1.44)\n\Rightarrow v_y=10.08\ m/s$

Net velocity is

$\Rightarrow v=√(10.92^2+10.08^2)\n\Rightarrow v=√(220.85)\n\Rightarrow v=14.86\ m/s$

Angle made is

$\Rightarrow \tan \theta =(10.08)/(10.92)\n\n\Rightarrow \tan \theta =0.92307\n\n\Rightarrow \theta =42.7^(\circ)$

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of 1.36 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 122 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature

Answers

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$