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Alexxx [7]
3 years ago
8

Four seismometers stations are the minimum needed to calculate the epicenter of an earthquake.

Physics
1 answer:
Gelneren [198K]3 years ago
4 0

<u>Answer:</u>

False, Four seismometer statitons are no required to find the earthquake's epicenter.

<u>Explanation:</u>

The least number of seismometer stations sufficient for finding the epicenter of earthquake is three. Epicenter is point on earth's crust from where the seismic rupture gets originated. Using the distances to the epicenter from atleast three seismometers the epicenter of earthquake can be determined. This distance is used as the radius and 3 circles are drawn. The point at which the 3 circles intersect is called as epicenter of the earthquake. With one seismometer station, we can determine only the distance to the epicenter. With two seismometer station, measurements are ambiguous. Using more than three seismometer stations produces redundant information. Hence three seismometer is sufficient to determine the epicenter of earthquake.

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Determine the magnitude and direction of the resultant force of the following free body diagram.
Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force (\vec F), in newtons, by vectorial sum:

\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j} (1)

\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}

\|\vec F\| \approx 599.923\,N

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)

Where \theta is the direction of the resultant force, in sexagesimal degrees.

\theta \approx 36.405^{\circ}

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0
2 years ago
Can someone help me with physics;((
vladimir2022 [97]

Answer: vf= 51 m/s and d= 112 m

Explanation: solution attached

4 0
3 years ago
A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Anastaziya [24]

Answer:

The gauge pressure is  P_g  =  2058 \ P_a

Explanation:

From the question we are told that

       The height of the water contained is  h_w  =  30 \ cm  =  0.3 \ m

        The height of liquid in the cylinder is  h_t  =  40 \ cm  = 0.4 \ m

       

At the bottom of the cylinder the gauge pressure is  mathematically represented as

        P_g  =  P_w + P_o

Where  P_w is the pressure of water which is mathematically represented as

      P_w  =  \rho_w  *  g * h_w

Now  \rho_w is the density of water with a constant values of  \rho_w  = 1000 \ kg /m^3

   substituting values

      P_w  = 1000 *  9.8 *  0.3

     P_w  =  2940 \  Pa

While P_o is the pressure of oil which is mathematically represented as

          P_o  =  \rho_o *  g *  (h_t -h_w )

Where \rho _o is the density of oil with a constant value

         \rho _o  = 900 \ kg / m^3

substituting values

       P_o  =  900 *  9.8 * (0.4 - 0.3)

       P_o  =  882 \ Pa

Therefore

      P_g  =  2940 - 882

      P_g  =  2058 \ P_a

6 0
3 years ago
An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600
dmitriy555 [2]

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

4 0
3 years ago
The first scientist to show that atoms emit any negative particles was
densk [106]
The first scientist to show that atoms emit any negative particles was : J.J Thomson
4 0
3 years ago
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