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jekas [21]
4 years ago
5

An electric charge q of mass m in an oscillating electric field Eosinot experiences force q Eosinot. Suppose it starts from rest

at time to, and position x-0. a) What is its velocity as a function of time? b) What is its kinetic energy as a function of time? c) What is the particle's time averaged kinetic energy (averaged over an integer # of complete cycles)? d) The vector potential A is defined so that the electric field is E at and the canonical momentum is then given by Pcan P can (t). mv + qA. Find A(t) and
Physics
1 answer:
Masja [62]4 years ago
7 0

Answer:

speed of the charge electric is  v = - (Eo q/m) cos t

Explanation:

The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,

          F = q Eo sint

a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity

        F = ma

        q Eo sint = ma

        a = Eo q / m sint

        a = dv / dt

        dv = adt

        ∫ dv = ∫ a dt

        v-vo = I (Eoq / m) sin  t dt

        v- vo = Eo q / m (-cos t)

We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0

        v = - (Eo q / m) cos t

b) Kinetic energy

       

         K = ½ m v2

          K = ½ m (Eoq / m)²2 (sint)²

         K = ¹/₂  Eo² q² / m sin² t

c) The average kinetic energy over a period

          K = ½ m v2

         <v2> = (Eoq / m) 2 <cos2 t>

The average of cos2 t = ½, substitute and calculate

          K = ½ m (Eoq / m)²  ½

          K = ¼ Eo² q² / m

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4. Are the forces acting on the chandelier unbalanced?​
kakasveta [241]

Answer:

If it is not an object in motion, all forces are balanced.

3 0
3 years ago
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the
Juli2301 [7.4K]

Answer:

4.47 km

Explanation:

If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.

From Pythagoras theorem

AD² = AE² + ED²

AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km

Hence, the van is 4.47 km from its initial point

3 0
3 years ago
A(n) __________ is a recording of a motion picture, or television program for playing through a television.
Soloha48 [4]

Answer:

Video

Explanation:

Hope this helps! If it does, drop a 5 star!

4 0
3 years ago
Read 2 more answers
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
3 years ago
If the mass of a moving object is doubled, which variable will also double? *
iVinArrow [24]

Answer:

potential energy

Explanation:

this is because potential energy= mgh

so when m= 2m

p.e. = 2 times the previous value

Therefore , When mass is doubled , Potential energy gets doubled. When mass is halved , Potential energy decreases by half.

4 0
3 years ago
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