Ask question

Ask question

All categories

2 years ago

11

The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff

ected. A state had 424 earthquakes over a given time span and a land area of 71,300 square miles. What is the seismic activity density for this state? Round to the nearest ten thousandth, if necessary.

2 answers:

Natalija [7]2 years ago

8 0

Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

$\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}$

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

$\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059$

The seismic activity density is 0.0059

Maurinko [17]2 years ago

8 0

Answer:

Seismic activity density will be 0.0059

Explanation:

We have number of earthquakes = 424

And area of the land 71300 square miles

We have to find the seismic activity density

It is given that seismic activity is the ratio of number of earthquakes to the affected land area

So seismic activity density $=\frac{424}{71300}=0.0059$

So the seismic activity density 0.00059

You might be interested in

Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.

WARRIOR [948]

Answer:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

And as we can see we have that:

$V_A = \frac{1}{4} V_B$

So then the best answer would be:

a. vA = vB/4

Explanation:

For this case we know the following conditions:

$L_A = L_B =L$ same length

$I_A = I_B =I$ both wires with the same current

Both wires are made of he same material, so then the number of electrons per cubic meter (n) are the same for both wires $n_A = n_B =n$

We also know that $r_A = 2 r_B$ where r represent the radius.

Since we know that a wire have a cylindrical form we can find the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

So then we have that $A_A = 4 A_B$

Now we know that from the definition the drift velocity of electron in a wire is given by:

$v_d = \frac{I}{neA}$

Where I is the current, n the number of electrons per cubic meter, e is the charge for the electron and A the area.

If we replace we have this:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

And as we can see we have that:

$V_A = \frac{1}{4} V_B$

So then the best answer would be:

a. vA = vB/4

6 0

2 years ago

A battery with internal resistance r is connected to a load resistance R. If R is increased, does the terminal voltage of the ba

Vitek1552 [10]

Answer:

The terminal voltage of the battery decreases.

Explanation:

An idea battery does not have internal resistance but a real practical battery always have an internal resistance r connected in series with the battery.

This internal resistance causes a voltage drop when load R is connected, a current I flows in the circuit which causes a voltage drop I*r in the battery therefore, the terminal voltage of the battery will decrease.

8 0

2 years ago

You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

Readme [11.4K]

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

$d_1 = v*t_1$

$d_1 = 10*20 = 200 m$

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

$d_2 = v*t = 20*20 = 400 m$

so the net displacement is given as

$\vec d = \vec d_2 - \vec d_1$

$\vec d = 400 - 200 = 200 m$

so it will be displaced by total displacement 200 m

8 0

2 years ago

A brick is resting on a rough incline as shown in the figure. The friction force acting on the brick, along the incline, is

Tasya [4]

B) equal to the gravitational force of the brick

6 0

2 years ago

Read 2 more answers

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

2 years ago