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Sholpan [36]
3 years ago
5

What difficulty will you encounter if you only have data from two recording station?

Physics
2 answers:
denpristay [2]3 years ago
6 0

<span>If you have only two data from two recording stations then you will be having a hard time finding the correct location of the epicenter. This is because triangulation method requires 3 recording station. If you have 2 recording station, the 2 circles will intersect at 2 points giving you 2 locations that could possibly be the epicenter.</span>

Zarrin [17]3 years ago
3 0

Answer:

The seismic data are often recorded using seismograph, that records the arrival of the P and S waves during an earthquake. This seismograph is used to locate the epicenter of an earthquake. During an earthquake, the seismographs placed in the nearby regions records the data regarding these seismic waves, and generates a circle.  

When three seismographs placed at three different locations forms three circles due to an earthquake, it intersect at a common point. This intersecting point is the approximate location of the epicenter. It is a very much efficient and are commonly used.

But when there are only two recording stations, then it becomes difficult to determine the epicenter because two seismographs at two different location can form two circles after recording the data and this two circle will intersect each other at two different point. As a result of which the difficulties will arise in locating the epicenter.

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Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib
Effectus [21]

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

6 0
2 years ago
Narysuj wykres zależności szybkości od czasu i drogi od czasu jeśli ciało porusza się ruchem jednostajnym z szybkością 45 m/s.
murzikaleks [220]
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7 0
3 years ago
The conversion of sugar to energy in the presence of oxygen is
Ahat [919]

Answer:

A. respiration.

Explanation:

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Additionally, mitochondria provides all the energy required in the cell by transforming energy forms through series of chemical reactions; breaking down of glucose into Adenosine Triphosphate (ATP) used for providing energy for cellular activities in the body of living organisms.

Basically, oxygen goes into the body of a living organism such as plants, humans and animals when they breathe while glucose is absorbed by the body when they eat.

Hence, the conversion of sugar to energy in the presence of oxygen is respiration.

8 0
3 years ago
A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
4 years ago
A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant
Juliette [100K]
<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force F applied to it, <u>as long as the spring is not permanently deformed</u>:

F=k (x-x_{o})    (1)

Where:

k is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

x_{o} is the length of the spring without applying force.

x is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth W of the block:

W=m.g   (2)

Where m=250g=0.25kg is the mass of the block and g=9.8\frac{m}{s^{2}}  is the gravity acceleration.

W=(0.25kg)(9.8\frac{m}{s^{2}})   (3)

W=2.45N   (4)

Knowing the force applied W and x=18cm=0.18m and x_{o}=0, we can substitute the values in equation (1) and find k:

W=k (x-x_{o})    (5)

2.45N=k (0.18m-0m)    (6)

<u>Finally:</u>

k=13.61\frac{N}{m}  

4 0
3 years ago
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