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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Vx = 2*cos30 = 1.73m's
Other words:
D = Vo*t = 1.4 * 21 = 31.5m. @ 30 Deg.
Dy = 31.5*sin30 = 15.75 m.
Answer:
See the explanation below.
Explanation:
The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.
s = displacement [m]
v and u = velocity [m/s]
t = time [s]
Now using these units in the given equation.
So the expression is good, and dimensional has consistency.
Answer:
it would
Explanation:
this is because as something rubs against a surface fast, friction heats it up, like when you rub a surface fast and it gets warm on the palm of your hand, that is friction. hope it helped :)
Answer:
W = 28226.88 N
Explanation:
Given,
Mass of the satellite, m = 5832 Kg
Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m
The time period of the orbit, T = 1.9 h
= 6840 s
The radius of the planet, R = 4.38 x 10⁶ m
The time period of the satellite is given by the formula
second
Squaring the terms and solving it for 'g'
g = 4 π² m/s²
Substituting the values in the above equation
g = 4 π²
g = 4.84 m/s²
Therefore, the weight
w = m x g newton
= 5832 Kg x 4.84 m/s²
= 28226.88 N
Hence, the weight of the satellite at the surface, W = 28226.88 N