All categories

4 years ago

Name the two scales that are used to measure earth intensity

Physics

1 answer:

nikklg [1K]4 years ago

3 0

Answer:

Although several scales have been developed over the years, the two commonly used today in the United States are the moment magnitude scale, which measures magnitude (M), or size, and the Modified Mercalli scale, which measures intensity.

You might be interested in

The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 {\rm rpm} .

damaskus [11]

Answer:

5833.33

Explanation:

$\alpha$ = Angular acceleration

$\theta$ = Number of revolutions

$\omega_i$ = Initial angular speed = 0

t = Time taken = 2 s

Final angular speed

$\omega_f=\dfrac{350000}{60}=5833.33\ rps$

From the equation of rotational motion we have

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{5833.33-0}{2}\\\Rightarrow \alpha=2916.665\ rev/s^2$

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2916.665\times 2^2\\\Rightarrow \theta=5833.33\ rev$

The number of revolutions is 5833.33

6 0

3 years ago

Earth’s atmosphere is in hydrostatic equilibrium. What this means is that the pressure at any point in the atmosphere must be hi

Misha Larkins [42]

Answer: The pressure that one experiences on the Mount Everest will be different from the one, in a classroom. It is because pressure and height are inversely proportional to each other. This means that as we move up, the height keeps on increasing but the pressure will keep on decreasing. This is the case that will be observed when one stands on the Mount Everest as the pressure is comparatively much lower there.

It is because as we move up, the amount of air molecules keeps on decreasing but all of the air molecules are concentrated on the lower part of the atmosphere or on the earth's surface.

Thus a person in a low altitude inside a classroom will experience high pressure and a person standing on the Mount Everest will experience low pressure.

6 0

4 years ago

A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe

Bogdan [553]

Answer:

A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

a_c = v² / r

the angular and linear variables are related

v = w r

we substitute

a_c = w² r

let's reduce the magnitudes to the SI system

w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

a_c = 418.88² 0.1

a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

v² = v₀² - 2g (y- y₀)

v² = 0 - 2 9.8 (0 + 1)

v =√19.6

v = -4.427 m / s

now let's look for the applied steel to stop the test tube

v_f = v + a t

0 = v + at

a = -v / t

a = 4.427 / 0.001

a = 4.43 10³ m / s²

4 0

3 years ago

What is a mixture?

Kazeer [188]

Potassium and Chlorine is a mixture.

8 0

3 years ago

Read 2 more answers

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial:

$0.8\cdot t^{2}-56.92\cdot t +3.872 = 0$ (3)

$t_{1} \approx 71.082\,s$ , $t_{2}\approx 0.068$

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

3 0

3 years ago