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Ivahew [28]
3 years ago
5

A total charge Q is distributed uniformly over a large flat insulating surface of area A . If the electric field magnitude is eq

ual to 1000 NC/ at a point located a perpendicular distance of 0.1 m away from the center of the sheet, then the electric field at a point a perpendicular distance 0.2 m from the center of the sheet is:_______
a. 1000N/C
b. 500N/C
c. Impossible to say since we are not given Q and A
d. 250 N/C
Physics
1 answer:
Andrew [12]3 years ago
3 0

Answer:

a. 1000N/C

Explanation:

Data mentioned in the question

Electrical field magnitude = 1000 NC

Perpendicular distance = 0.1 m

Perpendicular distance = 0.2 m

Based on the above information, the electric field is

As we know that

E = \frac{\sigma}{2\times E_o}

where,

\sigma = surface charge density

E = distance from nearby point to sheet i.e be independent

The distance at 0.1 and 0.2, the electric field would remain the same

So,

Based on the above explanation, the first option is correct

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Assuming a roughly spherical shape and a density of 4000 kg/m3, estimate the diameter of an asteroid having the average mass of
vredina [299]

Explanation:

It is given that,

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Mass of asteroid, m=10^{17}\ kg

We need to find the diameter of the asteroid. The formula of density is given by:

\rho=\dfrac{m}{V}

V is the volume of spherical shaped asteroid, V=\dfrac{4}{3}\pi r^3

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r = 2441311.12 m

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d = 4882622.24 m

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d=4.88\times 10^6\ m

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8 0
3 years ago
Un hamster esta sentado sobre un tocadisco cuya rapidez angular es constante si el hamster se mueve a un punto localizado al dob
kotegsom [21]

Answer:

b) se duplica

Explanation:

The disk is moving with constant angular velocity, let's call it \omega.

The linear velocity of a point on the disk is given by

v=\omega r

where r is the distance of the point from the axis of rotation.

In this problem, the object is moved at a distance twice as far as the initial point, so

r' = 2r

Therefore, the new linear velocity is

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So, the velocity has doubled, and the correct answer is

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8 0
2 years ago
A 21 g ball is shot from a spring gun whose spring has a force constant of 579 N/m. The spring can be compressed 1 cm. How high
vodka [1.7K]

Answer:

The answer  to the question is

The ball will go 0.14 meters high if the gun is aimed vertically

Explanation:

The energy in the spring → Energy, E = \frac{1}{2}kx^2

Where E = energy in the spring

k = Spring constant

x = Spring compression or stretch

Therefore E = \frac{1}{2} 579*0.01^2 = 0.02895 J

The spring energy is transferred  to the ball as kinetic energy based on the first law of thermodynamics which states that energy is neither created nor destroyed

Kinetic energy = KE = \frac{1}{2}mv^2

From which v = \sqrt{\frac{2*KE}{m} } = \sqrt{\frac{2*0.02895}{0.021} } = 1.66 m/s

from v² =u² - 2·a·S

Where v = final velocity = 0 m/s

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S = height

Therefore 0 = 1.66² - 2×9.81×S

or S = 1.66² ÷ (2×9.81) = 0.14 m

7 0
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