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sladkih [1.3K]
3 years ago
9

Suppose we were to attempt to use a similar machine to measure the charge-to-mass ratio of protons, instead. Suppose, for simpli

city, that we can get a "source" for a beam of protons as we do here for electrons (with a similar setup). What complications (if any) do you expect?
Physics
1 answer:
Igoryamba3 years ago
3 0

Answer:

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

Explanation:

The machine to measure the ratio of charge / mass of the electron, has two parts: a part where it accelerates the electrons in an electric field and a second section to charter the beam and measure its radius of curvature calculated from here the q / m ratio

In the case of having protons, the charge has the same value as that of the electrons, but with a positive charge, so the polarities of the fields should be changed.

The mass of the protons is much greater than the mass of the electrons, which introduced a significant difference in the excrement, since similar electric fields the speed of the protons is much less than the speed of the electrons, so the magnetic field through which the voters pass to have equivalent deflations in many cases this small values ​​of the magnetic field are not desirable due to the interference of the Earth's magnetic field.

Another solution could be to increase the electric field to have the protons with speeds similar to the electors, this possibility is not easy either, because the field of trunking of more than 5000 V would be needed, which are very dangerous.

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

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A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil h
tresset_1 [31]

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

Diameter = 25 cm

Magnetic field = 1.0 mT

Number of turns = 100

We need to calculate the current

Using the formula of magnetic field

B =\dfrac{\mu_{0}NI}{2\pi r}

I=\dfrac{B\times2\pi r}{\mu N}

Where, N = number of turns

r = radius

I = current

Put the value into the formula

I=\dfrac{1.0\times10^{-3}\times2\pi\times12.5\times10^{-2}}{4\pi\times10^{-7}100}

I=6.25\ A

Hence, The current passes through the coil is 6.25 A

6 0
3 years ago
A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0
4 years ago
Which of the following is not an example of kinetic energy being converted to potential energy?
KengaRu [80]

The list of choices you provided with your question
is utterly devoid of any such examples.

6 0
4 years ago
Read 2 more answers
Which phrase describes an atom?
kiruha [24]

You haven't included the list of choices that goes with the question, so it's
impossible for me to choose the correct one, or to help you choose it.

Regarding my ability to answer the question and collect the 5-point bounty,
I'm free to make up any phrase of my own that correctly describes an atom.

-- very very very very very very very tiny

-- includes even tinier particles, with electric charges
   both positive and negative

-- smaller than the wavelength of visible light

4 0
3 years ago
An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SCORPION-xisa [38]

Answer: 6.175 km

Explanation:

from the question, we have the following

velocity of the automobile = 95 km/g

velocity of the train = 75 km/h

length of the train = 1.30 km

since the automobile and the train are moving in the same direction, we need to find the velocity of the car relative to the train which will be their difference in speed = 95 - 75 = 20 km/h

we need to find the time it takes the automobile to overtake the train using the formula time = distance / speed , with the distance being the length of the train.

time (t) = 1.3 / 20

= 0.065 hour

now we can find the distance traveled by the automobile using the the time taken for it to overtake the train and the speed of the automobile.

therefore, distance = speed x time

     distance = 95 x 0.065 =6.175 km

5 0
4 years ago
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