All categories

2 years ago

Explain the challenges in developing an accurate rating system for earthquakes. What kinds of variables are there?

Physics

1 answer:

Lelu [443]2 years ago

6 0

Answer:

In turn, the main factors affecting earthquake shaking intensity are earthquake depth, proximity to the fault, the underlying soil, and building characteristics—particularly height. Let's take a look at the latter two (soil and buildings) and how they interact.

You might be interested in

The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this

svlad2 [7]

Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

Explanation: On Edgenuity!!

7 0

2 years ago

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago

Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi

morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = $\dfrac{0.029158}{4.03176}$ = 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0

2 years ago

Read 2 more answers

What statements accurately describe sunspots? Check all that apply.

polet [3.4K]

Answer:

sunspots are storms on the Suns surface

Sunspots are marked by intense magnetic activity

Sunspots produce solar flares and hot gassy ejections.

Sunspots can affect Earth’s climate.

Explanation:

I just did this lesson

6 0

2 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li

Elodia [21]

Answer:

a) E=228391.8 N/C

b) E=-59345.91N/C

Explanation:

You can use Gauss law to find the net electric field produced by both line of charges.

$\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}$

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

$E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})$

a) for y=0.200m, r1=0.200m and r2=0.200m:

$E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C$

b) for y=0.600m, r1=0.600m, r2=0.200m:

$E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C$

5 0

2 years ago