Ask question

Ask question

All categories

3 years ago

14

Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the

wire between them is cut, and the charged rod is taken away. Which one of the following is true? a. The spheres will attract each other. b. The spheres will repel each other. c. There will be no electrostatic force between the spheres

1 answer:

dangina [55]3 years ago

4 0

Answer:

a. The spheres will attract each other.

Explanation:

When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.

When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.

Hence the two spheres will experience some amount of electrostatic force between them.

You might be interested in

A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state

balu736 [363]

To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

$I = F*t$

Where,

F= Force

t= time

At the same time the moment can be described as a function of mass and velocity, that is

$P = m\Delta v \rightarrow P=m(v_1-v_2)$

Where,

m = mass

v = Velocity

From equilibrium the impulse is equal to the momentum, therefore

$I = p$

$Ft = m(v_1-v_2)$

PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be

$p=m(v_1-v_2)$

$p = 75*0.15$

$p = 1125Kg\cdot m/s$

Therefore the magnitude of the person's impulse is 1125Kg.m/s

PART B) From the equation obtained previously we have that the Force would be:

$Ft = m(v_1-v_2)$

$F(0.025)= 1125$

$F= 45000N$

Therefore the magnitude of the average force the airbag exerts on the person is 45000N

6 0

2 years ago

Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr

castortr0y [4]

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

$R = \frac{2.525}{1000}\times 50$

$R = 0.126 ohm$

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

$V = iR$

$V = 11 \times 0.126$

$V = 1.4 Volts$

so most appropriate answer in given options is

A. 1.8 Volts

8 0

2 years ago

In the equation 234Th90 -> 234Pa91 + X, which particle is represented by X?

taurus [48]

This process shows alpha decay so X represents an alpha particle. The characteristic of alpha decay is that the mass number stays constant while the proton number increases.

4 0

3 years ago

In the picture below, a doctor is having a video conference with a patient. What might be a positive effect of the technology be

GarryVolchara [31]

Answer:

Sorry to say but where is the photo???

The positive effect of technology being used might be using sethescope or checking BP rate if it is good.

Explanation:

6 0

3 years ago

What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic

Nata [24]

Answer:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, $\dfrac{dB}{dt}=3\ T/s$

Current, I = 6 A

Resistance of the solenoid, R = 17 ohms

We need to find the radius of the solenoid (r). We know that emf is given by :

$E=N\dfrac{d\phi}{dt}$

$E=N\dfrac{d(BA)}{dt}$

Since, E = IR

$IR=NA\dfrac{dB}{dt}$

$A=\dfrac{IR}{N.\dfrac{dB}{dt}}$

$A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}$

$A=0.188\ m^2$

or

$A=0.19\ m^2$

Area of circular cross section is, $A=\pi r^2$

$r=\sqrt{\dfrac{A}{\pi}}$

$r=\sqrt{\dfrac{0.19}{\pi}}$

r = 0.24 m

So, the radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.

5 0

3 years ago