A charge of +4q is placed at the center of an uncharged conducting shell. What will be the charges on the inner and outer surfaces of the
shell, respectively?
a) -29, +2q
b) -4q, +42
+42
c) -29, -29
d) +42q, -4q

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Ответ:

chelsearodi3985

26.08.2022

Physics

$\boxed{\sf Acceleration \ of \ train = 1 \ m/s^{2}}$

Given:

Initial speed (u) = 0 km/hr = 0 m/s

Final speed (v) = 36 km/hr

Time taken (t) = 10 sec

To Find:

Acceleration (a) of the train

Explanation:

$\sf 1 \ km/hr = \frac{5}{18} \ m/s \\ \therefore \\ \sf 36 \ km/hr = 36 \times \frac{5}{18} \ m/s \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \times \cancel{18} \times \frac{5}{ \cancel{18}} \ m/s \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \times 5 \ m/s \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \ m/s$

So,

Final speed (v) = 10 m/s

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ t: \\ \sf \implies 10 = 0 + a(10) \\ \sf \implies 10 = a(10) \\ \sf \implies a \times 10 = 10 \\ \\ \sf Dividing \ both \ sides \ by \ 10: \\ \sf \implies \frac{a \times 10}{\boxed{\sf 10}} = \frac{10}{\boxed{\sf 10}} \\ \\ \sf \frac{\cancel{10}}{\cancel{10}} = 1: \\ \sf \implies a = 1 \: m/s^2$

Acceleration of train = 1 m/s²

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A charge of +4q is placed at the center of an uncharged conducting shell. What will be the charges on the inner and outer surfaces of the shell, respectively? a) -29, +2q b) -4q, +42 +42 c) -29, -29 d) +42q, -4q