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Talja [164]
2 years ago
9

A man wishes to travel due north in order to cross a river 5 kilometers wide flowing due East at 3 kilometers per hour . if he c

an row at 10 kilometers per hour In still water, find either by scale drawing or calculation
1. the direction in which he must heads his boat in order to get to his destination directly opposite his starting point​
Physics
1 answer:
Liono4ka [1.6K]2 years ago
7 0

Answer:

17.46° W of N

Explanation:

His prow must point westward so that the westward component of his total velocity is equal to the eastward river speed.

3 = 10sinθ

sinθ = 3/10

θ = 17.4576... or about 17.46° W of N

it will take him 5 / 10cos17.46 = 0.52414... hrs. or about 31 min 27 s to cross.

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the first s-wave reaches a seismic station 22 minutes after an earthquake occurred. how long did it take the first p-wave to rea
Naddik [55]

The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

<h3>Time of travel of the P-wave</h3>

In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.

<h3>Relationship between speed and time</h3>

v ∝ 1/t

v₁t₁ = v₂t₂

t₁/t₂ = v₂/v₁

t₁/t₂ = 0.6v₁/v₁

t₁/t₂ =  0.6

t₁ = 0.6t₂

t₁ = 0.6 x 22 mins

t₁ = 13.2 mins

Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

Learn more about P-waves here: brainly.com/question/2552909

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7 0
2 years ago
Which of the following is a derived unit?<br><br> ampere<br> newton<br> second<br> kilogram
Soloha48 [4]

Newton is your answer.

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hope this helps

5 0
3 years ago
What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3
Olin [163]
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
4 0
3 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
3 years ago
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