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Rainbow [258]
3 years ago
14

P and s waves from an earthquake travel at different speeds, and this difference helps in locating the earthquake âepicenterâ (w

here the disturbance took place). (a) if an earthquake occurs and these two waves arrive at a particular seismic station with the p wave 2.0 min ahead of the s wave, then how far away was the earthquake? assume typical speeds of 8.5 km/s and 5.5 km/s for p and s waves, respectively.
Physics
1 answer:
mamaluj [8]3 years ago
8 0

Although the speeds are different, but the two waves will reach the same station so the same distance.

dp = ds

The formula for distance given velocity and time is:

d = v t

Substituting this to the 1st equation:

vp tp = vs ts

Since p wave arrived 2 min (120 s) earlier than s wave, therefore

tp = ts - 120

Then,

vp (ts – 120) = vs ts

8.5 (ts – 120) = 5.5 ts

8.5 ts - 1020 = 5.5 ts

3 ts = 1020

ts = 340 s

 

tp = ts – 120 = 340 – 120 = 220 s

 

Then using either dp or ds, we can calculate how far away the earthquake is:

dp = vp tp

dp = 8.5 * 220

dp = 1870 km = 1.9 * 10^3 km

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To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
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Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

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