A neutral metal ball is suspended by a string. A positively charged insulating rod is placed near the ball, which is observed to be attracted to the rod. This is because:____________. a. the ball becomes negatively charged by induction
b. the ball becomes positively charged by induction
c. the string is not a perfect insulator
d. there is a rearrangement of the electrons in the ball
e. the number of electrons in the ball is more than the number in the rod

Answers

Answer 1
Answer:

Answer:

d. there is a rearrangement of the electrons in the ball

Explanation:

Inside the neutral metal ball, there are equal no. of positive charges (protons) and negative charges (electrons). Normally, the charges are distributed evenly throughout the ball.

However, when the positively charged insulating rod is brought near, since positive charges and negative charges attract each other, the electrons (-ve charges) in the metal ball moves towards the side nearest to the rod. The metal ball gets attracted to the rod.

a and b are not correct because the rod is insulating, so electrons cannot be transferred between them to induce a net charge in the metal ball. the no. of electrons is unrelated to the attraction between opposite charges , so e is incorrect as well.


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The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant

Answers

Answer:

i believe that it is d

Explanation:

Final answer:

In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.

Explanation:

In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.

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The planet earth orbits around the sun and also spins around its own axis. 33% part (a) calculate the angular momentum of the earth in its orbit around the sun in kg • m2/s

Answers

The angular momentum can be found by the formula:
L = m · v · r

Where:
m = mass of Earth = 5.972×10²⁴ kg
v = velocity of Earth around the Sun = 2.978×10⁴ m/s
r = distance from Sun = 1.496×10¹¹ m

Now, apply the formula:
L = 5.972×10²⁴ · 2.978×10⁴ · 1.496×10¹¹ 
   = 2.661
×10⁴⁰ kg·m²/s

The angular momentum of Earth in its motion around the Sun is 2.661×10⁴⁰ kg·m²/s.

Is known:

m = Earth's mass = 5.972 × 10²⁴ kg

v = the speed of the Earth around the Sun = 2,978 × 10⁴ m / s

r = distance from the Sun = 1.496 × 10¹¹ m

Asked:

L?

Answer:

L = m · v · r

L = 5,972 × 10²⁴ · 2,978 × 10⁴ · 1,496 × 10¹¹

L = 2,661 × 10⁴⁰ kg · m² / s

So, the angular momentum of the Earth in its motion around the Sun is 2.661 × 10⁴⁰ kg · m² / s.

Further Explanation

In physics, momentum is a quantity related to the velocity and mass of an object. in classical mechanics, momentum (denoted by P) is defined as the product of mass and velocity, thus producing vectors.

The momentum of an object (P) having mass m and moving with velocity v is defined as:

P = mv

Mass is a scalar quantity, while velocity is a vector quantity. Multiplication of scalar quantities with vector quantities will produce vector quantities. So, momentum is a vector quantity. The momentum of a particle can be seen as a measure of the difficulty of settling an object. For example, a heavy truck has greater momentum than a light car that moves at the same speed. Greater force is needed to stop the truck compared to a lightweight car in a certain amount of time. (The magnitude of mv is sometimes expressed as linear momentum of the particle to distinguish it from angular momentum).

Speed ​​is a vector quantity that shows how fast an object is moving. The magnitude of this vector is named speed and is expressed in units of meters per second (m / s or ms − 1). The mass may be a property of an object that's accustomed to explain the varied behaviors of the item being monitored. In everyday use, mass is typically synonymous with weight. But in keeping with modern scientific understanding, the burden of an object is caused by the interaction of mass with the field

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Details

Grade: College

Subject: Physics

keywords: momentum

A 100 A current circulates around a 2.00-mm-diameter superconducting ring. A. What is the ring's magnetic dipole moment?
B. What is the on-axis magnetic field strength 4.70 cm from the ring?

Answers

Answer:

0.000314 Am²

6.049*10^-7 T

Explanation:

A

From the definitions of magnetic dipole moment, we can establish that

= , where

= the magnetic dipole moment in itself

= Current, 100 A

= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²

On solving, we have

= ,

= 100 * 3.14 * 0.000001

= 0.000314 Am²

B

=(0)/4* 2/³, where

(0) = constant of permeability = 1.256*10^-6

z = 4.7 cm = 0.047 m

B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]

B = 1*10^-7 * 0.000628/1.038*10^-4

B = 1*10^-7 * 6.049

B = 6.049*10^-7 T

Final answer:

The magnetic dipole moment of the superconducting ring is 0.000314 Ampere*m². The on-axis magnetic field strength 4.70 cm from the ring is 6.56 μT.

Explanation:

The magnetic dipole moment (μ) of a current (I) circulating around a loop of radius (r) is given by the formula μ = Iπr². Substituting the given values in SI units (I=100 Ampere, r=1.00 mm = 0.001 m), we get μ = 100 * π * (0.001)² = 0.000314 Ampere*m².

To find out the on-axis magnetic field strength (B) at a certain distance from the ring, we use the formula B = μ0/(4π) * (2μ/r³), where μ0 represents the permeability of free space. Plugging the values in SI units (μ0 = 4π  × 10-7 T*m/A, r=4.70 cm = 0.047 m), The magnetic field is calculated to be B = (4π  × 10-7 T*m/A) /(4π) * (2 * 0.000314  m² / 0.0473m³) = 6.56 × 10-6 T or 6.56 μT.

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A 1,200 kg car travels at 20 m/s. what is it’s momentum ?

Answers

The momentum of the car is  24000 Kg•m/s

Momentum is defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, we can obtain the momentum of the car as follow:

  • Mass = 1200 Kg
  • Velocity = 20 m/s
  • Momentum =?

Momentum = mass × velocity

Momentum = 1200 × 20

Momentum of car = 24000 Kg•m/s

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Answer:

24000 kg·m/s

Explanation:

Momentum is Mass x Velocity, so 1200 kg time 20 m/s =  24000 kg-ms/s

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.