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nataly862011 [7]
3 years ago
5

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

o say that the lines we use to represent an electric field indicate the direction in which a _______ test charge would initially move when released from rest. Which of the following fills in the three missing words correctly?
a. (positive; negative; negative)
b. (positive; negative; positive)
c. (negative; positive; negative)d. (negative; positive; positive)
Physics
1 answer:
lesantik [10]3 years ago
3 0

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

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The cosmic microwave background essentially looks the same in all directions. This is an example of.
pentagon [3]

In general, all views of the cosmic microwave background are identical. Isotropy is demonstrated by this.

<h3>What exactly is isotropy, for instance?</h3>

The Greek words isos (equal) and tropos, from which the term "isotropy" is derived, mean "uniform in all directions" (way). The material properties of anisotropic materials, such as graphite, differ depending on the direction, in contrast to isotropic materials like glass, which show the same properties in all directions.

There is no "centre" to an isotropic universe, which is another characteristic. The North and South Poles are produced by the rotation of the Earth, giving them a distinctive orientation, but the Universe is visible from every angle. When we think about the Big Bang, which is the origin of the Universe, this is a crucial point.

To know more about cosmic microwave background visit:

brainly.com/question/28197648

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3 0
1 year ago
(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.
DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

L=\dfrac{\mu_o N^2A}{l}

A is area

L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH

(b) The energy stored in the inductor is given by :

E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J

Hence, this is the required solution.

4 0
3 years ago
A satellite orbits the earth a distance of 2.50 × 10^7 m above the planet’s surface and takes 6.43 hours for each revolution abo
ASHA 777 [7]

Answer:

2.30 m /sec2

Explanation:

See the attached file;

7 0
4 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
Nuetrik [128]

Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
3 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
4 years ago
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