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Rudik [331]
3 years ago
13

Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical

surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 2.0 cm from the center of the two surfaces?
Physics
1 answer:
N76 [4]3 years ago
6 0

Answer:

E = K Q / R^2     by Gauss' Law where Q is the  charge enclosed by the surface of revolution and R is the distance from the enclosed charge

Since Q = d 4 pi r^2   where d is the charge density and r the radius of the inner sphere

E = K / R^2 * (4 d pi r^2) = 4 K pi d (r / R)^2 = 4 K pi d * 1/4

E =  9 * 10E9 * 3.14 * 40 * 10E-12 = 1.13 N / C

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What is the difference between a bound orbit and an unbound orbit around the sun?
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Answer:

Explanation:

The difference between a bound orbit and an unbound orbit around the sun is that:

An object on a bound orbit pursues the same way around the Sun again and again, while an object on an unbound orbit moves toward the Sun only a single time and afterward stays away forever &  never returns.

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Which is not part of nuclear chemistry? A. Strong nuclear force B. Double-replacement reactions C. Radioactivity D. Nuclear deca
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The correct choice is

B. Double-replacement reactions

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An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for
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3 years ago
A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init
anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

T=2\pi\sqrt{\frac{m}{K}}

T=2\pi\sqrt{\frac{0.3}{160}}

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

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3 years ago
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