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Shkiper50 [21]
3 years ago
14

Do earthquakes and volcanoes occur anywhere away from plate boundaries

Physics
1 answer:
aleksklad [387]3 years ago
8 0
No it doesn’t occur anywhere
There are areas that are prone to these earthquakes and volcanoes
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A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th
NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

v=r\omega

v = 4.50\times110\times\dfrac{2\pi}{60}

v=51.83\ m/s

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

8 0
2 years ago
What is a galvanic cell??
Kobotan [32]
Its an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell.
8 0
3 years ago
What is the relationship between lightning and atoms
Rufina [12.5K]

Answer:

The answer is A

Explanation:

Lightning is formed by electrons in the air

6 0
3 years ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min
Yuki888 [10]

Answer:

A. To find the mass flow rate.

We use= 220 x 0.355/ 60

= 1.3kg/s

B. Volume flowrate is = mass flowrate / density

But density is 1000kg/m³

= 1.3kg/s/ 1000kg/m³

= 0.0013m³/s

C. Flow speead at 1

= 0.0013m³/s / (2 x 10-2m)²

= 6.5m/s

D.flow speed at 2

0.0013m³/s / (8x 10-2m)²

=1.63m/s

E. Gauge pressure at point 1

= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)

= 119kpa

7 0
3 years ago
A block of mass 0.252 kg is placed on top of a light, vertical spring of force constant 4 825 N/m and pushed downward so that th
svp [43]

Answer:

10.77m

Explanation:

The elastic potential energy of the spring when compressed with the mass is converted to the kinetic energy of the mass when released. This ie expressed in the following equation;

\frac{1}{2}ke^2=\frac{1}{2}mu^2..............(1)

where k is the force constant of the spring, e is the compressed length, m is the mass of the block and u is the velocity with which the block leaves the spring after being released.

If we make u the subject of formula from equation (1) we obtain the following;

u=\sqrt{\frac{ke^2}{m}}................(2)

Given;

e = 0.105m,

k = 4825N/m,

m = 0.252kg,

u = ?

Substituting all values into equation (2) we obtain the following;

u=\sqrt{\frac{4825*0.105^2}{0.252}}................(2)\\u=14.53m/s

The maximum height attained is then obtained from the third equation of motion as follows, taking g as 9.8m/s^2

v^2=u^2-2gh.........(3)

v = 0m/s

Hence

h=\frac{u^2}{2g}\\h=\frac{14.53^2}{2*9.8}\\h= 10.77m

7 0
3 years ago
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