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3 years ago

The zone of earthquakes and volcanoes surrounding the pacific ocean is called

Physics

1 answer:

gulaghasi [49]3 years ago

3 0

Answer:

The Ring of Fire

Explanation:

The ring of fire is also called the Circum-Pacific Belt, it is a path along the pacific ocean consisting of active volcanoes and frequent earthquakes.

It has a length of approximately 40,000 kilometers. It lies on the edge of tectonic plates where the in-earth vibrations and geothermal energies are prone to erupt out.

Ring of fire inhibits about 75% o the earth's volcanoes and 95% of earthquakes occur in this region.

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A bowler who always left the same three pins standing could be considered a(n) ____ bowler.

RideAnS [48]

A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.

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3 years ago

the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su

CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ = 1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

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3 years ago

Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation

nikdorinn [45]

b. The sweater has a tendency to attract protons.

8 0

3 years ago

Read 2 more answers

The total resistance of a 15-ohm, an 65-ohm and a 35-ohm resistor connected in parallel

yanalaym [24]

Answer:

I think its 9.0397 Ohms

Explanation:

take the reciprocal of all the resistances: 1/15, 1/65, 1/35

then add them: = 151/1365

then reciprocal the answer: =1365/151

And chuck it on a calculator: =9.04 Ohms

I think this is right but I'm not entirely sure. Tell me if I'm right by the way!

3 0

3 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

3 years ago