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m_a_m_a [10]
2 years ago
8

Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m

ade up of many protons each with a kinetic energy of 3.25 x 10-15J. A proton has a mass of 1.673x10-27kg and a charge of 1.602x10-19C. What is the magnitude of a uniform electric field that will stop these protons in a distance of 2 m
Physics
1 answer:
Genrish500 [490]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 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

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It has to be used because you can not visually see them
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3 years ago
A solid, cylindrical wire conductor has radius R = 30 cm. The wire carries a current of 2.0 A which is uniformly distributed ove
blondinia [14]

Answer:

Explanation:

The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.

Applying Ampere's law :-

∫ B dl = μ₀ I      ; I is current passing through ampere's loop

B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2

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7 0
3 years ago
Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of
Illusion [34]

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

6 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

7 0
3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
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