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2 years ago

The mercalli scale is a scale from ________.

Physics

1 answer:

Mashcka [7]2 years ago

6 0

<h2>Answer: 1 to 12. </h2>

<u>The Mercalli seismological scale is a scale of 12 degrees</u>, where 1 is very weak and 12 catastrophic.

This scale owes its name to the Italian physicist Giuseppe Mercalli and was developed to assess the intensity of earthquakes through the effects and damages caused to different structures.

This means the <u>Mercalli scale measures how strong a earthquake has been by </u><u>its consequences and not by its magnitude</u><u>, therefore it is based on </u><u>empirical observations.</u>

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What is the hang time when the person moves 6 m horizontally during a 1.25 m high jump?

AlekseyPX

Answer:

1 sec

Explanation:

Horizontal distance (x) = 6m

Vertical distance (y) = 1.25m

Hang time is the duration the object is in the air before it reaches maximum height.

The time of free fall is given by

t = √2y/g

g = acceleration due to gravity

t = √(2*1.25)/9.8

t = √2.5/9.8

t = 0.5secs

Hang time = 2*0.5

= 1 sec

3 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co

sp2606 [1]

Answer: The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

$I = \sqrt{\frac{P}{R} }$

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

$I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }$

Then, simplify the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} }$

Rationalize the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}$

Simplify the numerator by finding its factors:

$I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}$

The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

8 0

3 years ago

A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (a) What is the distance from the

Zarrin [17]

The distance from the horizontal top surface of the cube to the water level is "6.282 cm".

<h3>What is Archimedes' principle?</h3>

According to Archimedes' principle, the weight of the fluid that the body displaces is equal to the upward buoyant force that is applied to a body submerged in a fluid, whether fully or partially. The Archimedes' principle is a fundamental physical law in fluid mechanics. It was created by Syracuse's Archimedes.

According to Archimedes' principle, a body submerged in a fluid experiences an upward force proportional to the weight of the fluid that has been displaced. One of the prerequisites for equilibrium is this. We believe that the buoyancy force, also known as the centre of buoyancy, is situated in the middle of the submerged hull.

From Archimedes' principle, we get

$\rightarrow L^3 \rho_{\text {Wood }} &=L^2 d \rho_{\text {Water }} \\$