The angular velocity of the propeller is 136.1 rad/s; linear velocity is 153.1 m/s; centripetal acceleration 20835.2 m/s² and 2123.8 g.
<h3>What is the angular velocity of the propeller?</h3>
The angular velocity of the propeller in rad/s is given as follows:
1300 rev/min = 1300 * 2π/60 = 136.1 rad/s.
b. The linear velocity, v = radius * angular velocity
Linear velocity, v = 2.25/2 * 136.1
v = 153.1 m/s
c. Centripetal acceleration, 

Centripetal acceleration in terms of g; 
Therefore, the angular velocity of the propeller is 136.1 rad/s; linear velocity is 153.1 m/s; centripetal acceleration 20835.2 m/s² and 2123.8 g.
Learn more about angular velocity and centripetal acceleration at: brainly.com/question/10703948
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Answer:
276.62 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)
Equation of motion

<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s
Heat<span> capacity ( C ) </span>does change with mass<span>. However, </span>specific heat<span> is the </span>heat<span>capacity per unit </span>mass<span> ( c=Cm ). Therefore if you double the amount of </span>mass<span> in your system, you've doubled its </span>heat<span> capacity, but you've kept the </span>specific heat<span> the same. ... </span>Specific<span> gravity is another such quantity.</span>
<u>Answer:</u>
Option A is the correct answer.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
First walking 1.2 km north, displacement = 1.2 j km
Secondly 1.6 km east, displacement = 1.6 i km
Total displacement = (1.6 i + 1.2 j) km
Magnitude = 
Angle of resultant with positive X - axis =
= 36.87⁰ east of north.