Ask question

Ask question

All categories

I am Lyosha [343]

3 years ago

5

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

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.

You might be interested in

Usimov [2.4K]

Answer:

A

Explanation:

7 0

2 years ago

Read 2 more answers

A sharp raises a note by

vivado [14]

The answer should be B. A half step

8 0

3 years ago

Read 2 more answers

If you good with science multiple choice

miss Akunina [59]

Answer:

If it had more or less mass, the atmosphere would be very different with either too much ammonia and methane or too little oxygen and water

Explanation:

6 0

2 years ago

21. A toy car starts from rest and begins to

Bond [772]

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

8 0

3 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

5 0

3 years ago