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kykrilka
1 day ago
15

A clock has an hour hand of length 2.4 cm and a minute hand of length 3.8 cm. (a) Calculate the position and velocity of the hou

r hand at noon. (b) Calculate the position and velocity of the minute hand at 12:15.
Physics
1 answer:
Ostrovityanka [942]1 day ago
6 0

Answer:

  1. At 12:00, the hour hand is positioned at 0 degrees, with its tip moving at an estimated 1.26 cm/hour.
  2. At 12:15, the minute hand is at 90 degrees, traveling at roughly 23.88 cm/hour.

Explanation:

The position is established by an angle, which begins at the 12:00 point and extends clockwise.

  1. At noon, the hour hand is at the 12:00 position, thus its position is 0 degrees; it has an angular velocity of (2π)/(12 hours) = π/(6 hours). Therefore, the tip moves at a linear velocity of (2.4 cm) * π/(6 hours) ≅ 1.26 cm/hour.
  1. At 12:15, the minute hand is at the 3 o'clock position, giving it a position of 90 degrees or π/2; its angular velocity is (2π)/(1 hour) = 2π/hour. Thus, the tip moves at a linear velocity of (3.8 cm) * 2π/hour ≅ 23.88 cm/hour.
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Answer:

F=126339.5N

Explanation:

To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:

Fw=W+Fi+F

where

Fw=   force or weight exerted by the water column above the submarine.

To calculate Fw, we can use:

Fw=h. γ. A

h=height

γ= specific weight of seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w represents the hatch weight = 200N

Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:

Fi=PA=101325x0.7=70927.5N

Finally, the force needed to open the hatch is determined by the original equation:

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N

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