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evablogger [386]
3 years ago
7

A proton is traveling horizontally to the right at 4.60×106 m/s . Part A Find (a)the magnitude and (b) direction of the weakest

electric field that can bring the proton uniformly to rest over a distance of 3.50 cm . Part C How much time does it take the proton to stop after entering the field? Part D What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?
Physics
1 answer:
Feliz [49]3 years ago
3 0

Answer:

Explanation:

u = 4.6 x 10^6 m/s

Let E be the electric field

s = 3.5 cm = 0.035 m

v = 0

a = qE / m

So use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (1.67 x 10^-27 x 0.035)

E = 3865 N/C

(a) The magnitude of electric field is 3865 N/C

(b) the direction of electric field is opposite to the direction of motion of proton, i.e., towards left.

(c) Let t be the time taken

v = u + a t

0 = 4.6 x 10^6 - (1.6 x 10^-19 x 3865) t / (1.67 x 10^-27)

t  = 1.24 x 10^-5 sec

(d) For electron, the direction of electric field is same the direction of electron, i.e., rightwards.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (9.1 x 10^-31 x 0.035)

E = 2.1 N/C

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Answer:

112.17 m/s

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Explanation:

h = 3.18 x 10^10 m

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Gate-control theory

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A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
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Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

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