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3 years ago

wink is the seismic station closest to the earthquake. describe two evidences from the seismogram that demonstrates this?

Physics

1 answer:

kotykmax [81]3 years ago

5 0

Well, we don't know if it's the CLOSEST one, but it's closer than the other ones on this print-out.

There are actually three clues to this in the seismograms:

1). The waves from the tremor reach the Wink station first ... the P wave gets there 50 seconds before Chicago and 3 whole minutes before Tampa.

2). They are stronger (greater amplitude) at Wink than at Chicago or Tampa.

3). The delay between the P wave and the S wave is the smallest at Wink. This delay between the two waves grows as the detector is farther from the quake.

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Why is there more potential for wind energy in the upper himalayan in nepal?

exis [7]

Answer:

<em><u>The potential area of wind power in the country is about 6074 sq. km with wind power density greater than 300 watt/m2. More than 3,000 MW of electricity could be generated at 5 MW per sq km. The commercially viable wind potential of the country is estimated to be only about 448 MW.</u></em>

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3 years ago

You are loading a refrigerator weighing 2267 N onto a truck, using a wheeled cart. The refrigerator is raised 1.09 m to the truc

meriva

Answer:

a).

Wmin= 2471.03 J

W=1603.01 J

Explanation:

Weight, w= 2267 N

$w= m*g\\W=m*g*h \\W=w*h$

Minimum work 'h' is the distance the refrigerator is raised h=1.09m

$W_{min}=2267 N* 1.09m\\W_{min}=2471.03 J$

The motion is no frictional force so, the magnitude of the force with a angle of 45.0° is find using:

$W=m*g*h'\\h'= sin(45)\\W=2267N*sin(45)\\W=1603.01 J$

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3 years ago

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

Svetradugi [14.3K]

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

$F = \frac{mv^2}{r}$

Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

$T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

$T = 16.8 s$

4 0

3 years ago

Read 2 more answers

suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in

ivolga24 [154]

Answer:

Half: 6 cm^2 Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

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3 years ago

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Has a skateboard. your gonna have to give more details the. that just one .

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3 years ago