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EastWind [94]
2 years ago
8

A particle beam is made up of many protons, each with a kinetic energy of 3.25times 10-15 J. A proton has a mass of 1.673 times

10-27 kg and a charge of +1.602 times 10-19 C. What is the magnitude of a uniform electric field that will stop these protons in a distance of 2 m?
Physics
1 answer:
ArbitrLikvidat [17]2 years ago
3 0

Answer:

The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x 10^{-4} N/C

Explanation:

given information,

kinetic energy, KE = 3.25 x 10^{-15} J

proton's mass, m = 1.673 x 10^{-27} kg

charge, q = 1.602 x 10^{-19} C

distance, d = 2 m

to find the electric field that will stop the proton, we can use the following equation:

E = F/q

   = (KE/d) / q ,        KE = Fd --> F = KE/d

   = KE/qd

    = (3.25 x 10^{-15} J) / (1.602 x 10^{-19} C)(2 m)

    = 1.01 x 10^{-4} N/C

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sergij07 [2.7K]

The two flaws in her experiment’s design are

<span>- She introduced at least one confounding variable.</span> <span>- She tried to test multiple hypotheses at a time</span>

 In the above mentioned experiment she had to have four samples to prove four hypotheses, each one separately and not to mix two hypotheses in an alone sample, that what it brings as consequence is the confusion.

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2 years ago
Select the correct answer from each drop-down menu.
bagirrra123 [75]

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

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Therefore, an object that is negatively charged has more electrons than protons.  An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

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A container, partially filled with water, is resting on a scale that measures its weight. Suppose you place a 200 g piece of woo
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an NHL hockey player (m=90 kg) stands motionless on the ice . the 10 kg stanley cup is thrown to the player at 4m/s. after he ca
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Answer:v=0.4 m/s

Explanation:

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Conserving momentum

m_1\cdot 0+m_2u=(m_1+m_2)v

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4 0
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How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Airida [17]

Answer:

See explanation

Explanation:

Solution:-

- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.

- The angle that each fringe forms on the screen is defined by ( θ ).

- The order of bright/dark spot is defined by an integer ( n )

- The wavelength of the incident light is ( λ )

- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The above given formulation is for constructive interference.

- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.

- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,

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- Substitute the above approximation in the given relation of diffraction gratings as follows:

                            y_n = n*lamda*L*N

- To double the distance between the screen and grating we will use the above relation with ( 2L ):

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Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.

- Similarly, the line density of the grating ( N ) was doubled. Then,

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Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.

4 0
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