1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
irakobra [83]
3 years ago
8

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

e surface?
Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer :

Velocity will be 3.28\times 10^{-11}m/sec

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere  = =1.6+\frac{1.5}{2}=2.35mm = 0.00235m

Force on electron is given by F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}, here q is charge on sphere and e is charge on electron

F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N

This force work as centripetal force

So F=\frac{mv^2}{r}

4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

You might be interested in
The movement of the liquid in a thermometer shows changes in temperature. An increase in temperature indicates the molecules in
ahrayia [7]

Answer:

The movement of the liquid in a thermometer shows changes in temperature. An increase in temperature indicates the molecules in the liquid *

Explanation:

7 0
3 years ago
A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may
12345 [234]

Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

Generally the volume of the rock is mathematically evaluated as

          V    =  \frac{m_r}{\rho}

substituting values

        V    =  \frac{5000}{4.3}

       V    =  1162.7 \  dm^3

The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

      W_r  =  m_r * g

     W_r  =  5.0 *  9.8

     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

       T  =  43.302 \ N

4 0
3 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
The earth's magnetic field deflects the flow of current from?
alexgriva [62]
I don't really know the answer but maybe north pole and south pole?
7 0
4 years ago
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and
Nimfa-mama [501]

Answer:

I think 9.5

Explanation:

............

6 0
3 years ago
Other questions:
  • Discuss renewable and nonrenewable resources. Give at least two advantages and two disadvantages of each. (4 points)
    15·1 answer
  • Describe how evaporation relates to heat regulation in your body
    15·1 answer
  • Communication with submerged submarines via radio waves is difficult because seawater is conductive and absorbs electromagnetic
    8·1 answer
  • Assume that on a physics diagram, you saw a croquet ball moving along the grass and above it was an arrow labeled "3.5 m/s" that
    8·1 answer
  • Questions to consider:
    12·1 answer
  • Volume (mL) Mass (g)
    6·2 answers
  • What is the CGS unit of current?
    10·2 answers
  • You need to move a 132 kg sofa to a different location in the room. It takes a force
    5·1 answer
  • A mason stretches a string between two points 70 ft apart on the same level with a tension of 20 lb at each end. If the string w
    11·1 answer
  • There are pros and cons to both e-readers and paper books. Do you think the pros outweigh the cons more for modern electronic re
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!