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Mamont248 [21]
3 years ago
9

An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg

e Q is brought close to, but not touching, sphere A. Describe what happens to the charges on A and B as they are brought close but not touching. If we now remove sphere B and place it far away, what is the charge on sphere A
Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

0

Explanation:

  • If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
  • As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:

       Q +(-Q) =0.

  • If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
  • The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.
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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
ivann1987 [24]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

\tau=F\times r

\tau=82.0\times0.150

\tau=12.3\ N-m

Now, The angular acceleration

\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}

\dfrac{d\omega}{dt}=20.73\ rad/s^2

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

\tau=I\times\dfrac{d\omega}{dt}

I=\dfrac{I}{\dfrac{d\omega}{dt}}

I=\dfrac{12.3}{20.73}

I= 0.593\ kg-m^2

Hence, The moment of inertia of the wheel is 0.593 kg-m².

5 0
4 years ago
This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ___________
melisa1 [442]

Answer:

have a component along the direction of motion that remains perpendicular to the direction of motion

Explanation:

In this exercise you are asked to enter which sentence is correct, let's start by writing Newton's second law.

circular movement

          F = m a

          a = v² / r

          F = m v²/R

where the force is perpendicular to the velocity, all the force is used to change the direction of the velocity

in linear motion

         F = m a

where the force is parallel to the acceleration of the body, the total force is used to change the modulus of the velocity

the correct answer is: have a component along the direction of motion that remains perpendicular to the direction of motion

8 0
3 years ago
What happens to particle spacing when temperature increase, What we call this process?
Slav-nsk [51]

Answer:

The atoms start vibrating really fast, making the space between them larger. It's an increase in Kinetic Energy.

6 0
4 years ago
If you can simply pour sand into a cup then why is it not a liquid?
sleet_krkn [62]
If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.
6 0
4 years ago
Read 2 more answers
A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distan
OleMash [197]

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Speed of the car, v = 100 km/h = 27.77 m/s

The coefficient of kinetic friction of the tires, \mu_k=0.5

Let f is the net force acting on the body due to frictional force, such that,

-f=ma

a=\dfrac{-f}{m}

a=\dfrac{-\mu _k mg}{m}

a=\mu_k g

a=-0.5\times 9.8

a=-4.9\ m/s^2

We know that the acceleration of the car in calculus is given by :

v.dv=a.dx, x is the stopping distance

\int\limits^0_v {v.dv}=\int\limits^x_0 {a.dx}

\dfrac{v^2}{2}|_v^0=ax

0-(27.77)^2=-2\times 4.9x

On solving the above equation, we get, x = 78.69 meters

So, the stopping distance for the car is 78.69 meters. Hence, this is the required solution.

6 0
3 years ago
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