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1 year ago

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How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that

the phrase "a great many" means that the total charge on the rod dominates any charge movement induced by the near presence of the charged ball.

1 answer:

sleet_krkn [62]1 year ago

6 0

Answer: IT IS STRONGLY REPELLED

Explanation: The laws of guiding magnetic attraction or repulsion of Magnetic materials,states that when like poles are brought together they repel each other, but when unlike poles are brought together they are attracted.

The rod will be strongly repelled because the forces on the rod is greater and has the same Polarity as the charged ball.

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The current in a long solenoid of radius 6 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of

jonny [76]

Answer:

53.63 μA

Explanation:

radius of solenoid, r = 6 cm

Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2

n = 17 turns / cm = 1700 /m

di / dt = 5 A/s

The magnetic field due to the solenoid is given by

B = μ0 n i

dB / dt = μ0 n di / dt

The rate of change in magnetic flux linked with the solenoid =

Area of coil x dB/dt

= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt

= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4

The induced emf is given by the rate of change in magnetic flux linked with the coil.

e = 2.145 x 10^-4 V

i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA

6 0

1 year ago

Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c

tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm= 2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

$\frac{k*q_{2} *q_{3} }{d_{2}^{2} } -\frac{k*q_{1} *q_{3} }{d_{1}^{2} }=0$

We eliminate k*q₃ of the equation

$\frac{q_{1} }{d_{1}^{2} } = \frac{q_{2} }{d_{2}^{2} }$

$q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2} }$

$q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4} }{5^{2}*10^{-4} }$

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0

1 year ago

A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0

Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

$vsin(\alpha) + gt = 0$

$t = \frac{-vsin(\alpha)}{g}$

So the vertical distance it travels within time t is

$y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}$

$y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}$

$y = \frac{-v^2sin^2(\alpha)}{2g}$

Similarly the horizontal distance it travels within time t is:

$x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}$

$x = \frac{-v^2sin(2\alpha)}{2g}$

We can pre-calcualte the quantity $\frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54$

So $y = 57.54sin^2(\alpha)$

$x = 57.54sin(2\alpha)$

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

8 0

1 year ago

Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac

dangina [55]

Explanation:

Relation between electric field and charge density is as follows.

E = $\frac{\sigma}{2 \epsilon}$

where, $\sigma$ = charge density

$\epsilon$ = permittivity of free space = $8.85 \times 10^{-12}$

So, $E_{\text{outside}}$ = 0

$E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}$

or, $E_{inside} = \frac{\sigma}{\epsilon}$

Now, formula to calculate the potential difference of two conductors is as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

It is given that,

d = 6.0 mm = $6 \times 10^{-3} m$

$\sigma = 40 \times 10^{-12} C/m^{2}$

Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

= $\frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}$

= 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

7 0

2 years ago

What is the mass of an object that weighs 686N on Earth?

Oxana [17]

Answer:

70 kg is the mass of the object

Explanation:

This question can be solved with this simple formula:

Weight force = mass . gravity

686 N = mass . 9.8 m/s²

686 N / 9.8 m/s² = mass → 70 kg

Note → 1N = 1 kg . m / s²

8 0

2 years ago