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g100num [7]
3 years ago
12

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, i

s produced by charges on the plates. The center plane at x = 0.00 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m. with an initial kinetic energy K = 300 eV, in the positive x-direction.
a. Determine the initial velocity of the electron at x = 0.0 m.
b. What is the potential difference between the plates?
c. At a certain point the electron stops momentarily and then reverses direction. What is the electric potential at this point?
d. What is the kinetic energy of the electron when it reaches plate A?
Physics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

b. E = V/d

c. V= 0 V (momentarily the pd changes to zero)

d KE= 300*3600 v = 1.08 MeV

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Answer:7 cm/s

Explanation:

Given

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\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

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14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

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2 years ago
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4 0
3 years ago
An object of mass 25kg is falling from the height h=10 m. calculate
r-ruslan [8.4K]

Answer:

a=2500J,b=1000K,c=1000J,d=14.142m/s

Explanation:

V²=U²+2gh

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potential energy=mgh

p.e=25×10×10=2500J

pe+ke=2500+2500=5KJ

b).mgh=25×10×4=1000J

c). V²=U²+2gh

V²=0+2×10×4

V²=80

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=(1/2)25×80

=1KJ

d). From my first paragraph V²=200

V=√200

V=14.142m/s

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