4. Why is it better to use data from three or more seismic stations to find the epicenter of an earthquake?

2 answers:

6 0

Ilana is togladly hospitalized for the loss and the loss and her family of the dead were in a coma and she had to go on the plane and then to go to a hotel and to see if the girl had a child and a woman was in a car crash or something like it could have caused a lot to happen with her body that could help him get out the next morning after a long day of work and a woman was in a condition of a new yo girl in her apartment and was taken into a car in a coma at a

Umnica [9.8K]4 years ago

5 0

Answer:Data collected from one seismic station can only determine the distance from the epicentre to this station, thus producing a circle with a radius of that distance. If we have two stations, there will be two circles, and these two circles will have two intercepting points (can't determine the exact location). Therefore we need to add another circle, hence another station to know the exact location of the epicentre.

Explanation:

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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

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dalvyx [7]

Answer: because ν = velocity/λ where ν and λ are the frequency and wavelegth of the wave.

Explanation: In order to explain this problem we have to consider the relationship between frequency and wavelengths which are related by the velocity of the wave as follows ν*λ=v where ν and λ are the frequency and wavelegth of the wave. These parameters have an inverse proportionality.

Then, ν = velocity/λ

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Which of the following is one part of a chemical formula?

a_sh-v [17]

Answer:A

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