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Vika [28.1K]
3 years ago
9

In separate experiments, a large number of particles (all with the same charge, but with a wide variety of masses and speeds) ar

e fired into a region containing a uniform magnetic field. The velocities of the particles are all perpendicular to the field. What do the particles that move in orbits of the same radius have in common
Physics
1 answer:
likoan [24]3 years ago
7 0

Answer:

  they have the same mass

Explanation:

The force applied by the field is a function of the charge and velocity, so the acceleration experienced by a particle will be dependent upon its mass. Particles in orbits with the same radius are exhibiting the same acceleration, so must have the same mass.

You might be interested in
A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,
Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

     F = m a

    a = F / m

We use kinematics to find lips

    v = v₀ + a t

    v = v₀ + (F / m) t

The moment is defined by

    p = m v

The moment change

    Δp = m v - m v₀

Let's replace the speeds in this equation

    Δp = m (v₀ + F / m t) - m v₀

    Δp = m v₀  + F t - m v₀  

    Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0
3 years ago
A. Write two or three sentences to describe the conductivity of an insulator. Explain its conductivity in terms of the electrons
aliya0001 [1]

PART A)

Conductivity of insulator is very small as there is no free electrons to conduct the current trough that medium

So here number of conduction electrons are very less in insulators

PART B)

Resistance is the property of a conducting medium which will oppose the flow of current trough it

Resistance of wire directly depends on its length so resistance of long wire will be more than the resistance of short wire

Resistance inversely depends on the area so if a wire has more crossectional area then its resistance must be small

PART C)

power of light bulb is defined as rate of electrical energy

it is given by formula

P = i V

here we know that

i = 1.46 A

V = 120 volts

so power is given as

P = 1.46 \times 120

P = 175.2 Watt

3 0
3 years ago
Read 2 more answers
Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift
sammy [17]

Answer:

The doppler effect equation is:

f' = \frac{v +v0}{v - vs}*f

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) -  (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0
3 years ago
Need help with this question
Alika [10]

Answer:

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Digital art

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Digital art is an artistic work or practice that uses digital technology as part of the creative or presentation process. Since the 1960s, various names have been used to describe the process, including computer art and multimedia art.[1] Digital art is itself placed under the larger umbrella term new media art.[2][3]

Maurizio Bolognini, Programmed Machines (Nice, France, 1992–97). An installation at the intersection of digital art and conceptual art (computers are programmed to generate flows of random images which nobody would see).

The image of the computer virus Chernobyl, created by Ukrainian new media artist Stepan Ryabchenko in 2011.

Irrational Geometrics digital art installation 2008 by Pascal Dombis

Joseph Nechvatal birth Of the viractual 2001 computer-robotic assisted acrylic on canvas

The Cave Automatic Virtual Environment at the University of Illinois, Chicago

After some initial resistance,[4] the impact of digital technology has transformed activities such as painting, drawing, sculpture and music/sound art, while new forms, such as net art, digital installation art, and virtual reality, have become recognized artistic practices.[5] More generally the term digital artist is used to describe an artist who makes use of digital technologies in the production of art. In an expanded sense, "digital art" is contemporary art that uses the methods of mass production or digital media.[6]

Lillian Schwartz's Comparison of Leonardo's self portrait and the Mona Lisa based on Schwartz's Mona Leo. An example of a collage of digitally manipulated photographs

The techniques of digital art are used extensively by the mainstream media in advertisements, and by film-makers to produce visual effects. Desktop publishing has had a huge impact on the publishing world, although that is more related to graphic design. Both digital and traditional artists use many sources of electronic information and programs to create their work.[7] Given the parallels between visual and musical arts, it is possible that general acceptance of the value of digital visual art will progress in much the same way as the increased acceptance of electronically produced music over the last three decades.[8]

Digital art can be purely computer-generated (such as fractals and algorithmic art) or taken from other sources, such as a scanned photograph or an image drawn using vector graphics software using a mouse or graphics tablet.[9] Though technically the term may be applied to art done using other media or processes and merely scanned in, it is usually reserved for art that has been non-trivially modified by a computing process (such as a computer program, microcontroller or any electronic system capable of interpreting an input to create an output); digitized text data and raw audio and video recordings are not usually considered digital art in themselves, but can be part of the larger project of computer art and information art.[10] Artworks are considered digital painting when created in similar fashion to non-digital paintings but using software on a computer platform and digitally outputting the

6 0
3 years ago
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