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STALIN [3.7K]
3 years ago
8

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

hat is the kinetic energy of this charge at x = 4m?
Physics
1 answer:
ANEK [815]3 years ago
8 0
The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
F=qE
where in our problem q=3 \mu C= 3 \cdot 10^{-6} C and E=2 kN/C=2000 N/C, so the force is
F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N

The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J
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Wave A has a longer wavelength than wave B, but their amplitudes are the same. Which carries more energy?
dangina [55]
The answer to this question would be B.
8 0
3 years ago
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Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m
denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0
3 years ago
Earth can be modeled with five layers based on the strengths and physical properties of the materials in them. Put the layers in
andre [41]

Answer:

Crust, Upper mantle, mantle, outer core, inner core

Explanation:

The Earth's layers have been clasified in 5 according to the materials that conform them, theri physical properties, strengths and also their state of matter. We all know how the outer layer of the Earth looks like, but if we start to dig a huge hole we are going to see different types por materials due to a change in pressure, temperature, and other factors. At the very center of the Earth there's what's called "core". The core is liquid and at extremely high temperatures. This is because of the enormous amount of pressure the rest of the Earth is putting it under. So, if we list the different layers of the Earth according to the materials they are made of, from the Earth's surface to the core, the answer is:

1) Crust (surface)

2) Upper Mantle

3) Mantle

4) Outer core

5) Inner core

In some books you may find a layer called Lithosphere. Tis layer consists not only of the crust, but also it contains the transition zone between the upper mantle and the crust.

4 0
3 years ago
Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up
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8 0
3 years ago
Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
3 years ago
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