Answer:
magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Explanation:
Given the data in the question;
Speed of carousel N = 24 rpm
From the diagram below, selected path direction defines the Axis of slip.
Hence, The Coriolis is acting along the axis of transmission
Now, we determine the angular speed ω of the carousel.
ω = 2πN / 60
we substitute in the value of N
ω = (2π × 24) / 60
ω = 2.5133 rad/s
Next, we convert the given velocity from mph to ft/s
we know that; 1 mph = 1.4667 ft/s
so
= 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s
Now, we determine the magnitude of the Coriolis acceleration
= 2( × ω )
we substitute
= 2( 8.8002 ft/s × 2.5133 rad/s )
= 44.235 ft/s²
Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Answer:
5 seconds
Explanation:
sound travels 1 km in roughly 3 secs so 1 mile in roughly 5 secs.
The answer is A 0 degrees is the freezing point of water
Plz mark brainliest
Answer:
not work
Explanation:
in a series circuit, everything meaning the electrons are flowing on one path, therefore, it wouldn continue to work.
Answer:
(A) 0.54 kg.m^{2}
(B) 0.0156 N
Explanation:
from the question you would notice that there are some missing details, using search engines you can find similar questions online here 'https://www.chegg.com/homework-help/questions-and-answers/small-ball-mass-120-kg-mounted-one-end-rod-0860-m-long-negligible-mass-system-rotates-hori-q7245149'
here is the complete question:
A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.
solution
mass of the ball (m) = 1.5 kg
length of the rod (L) = 0.6 m
angular velocity (ω) = 4900 rpm
air drag (F) = 2.60 x 10^{-2} N = 0.026 N
(take note that values from the original question are used, with the exception of the air drag which was not in the original question)
(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia =
where m = mass of ball and L = length of rod
= 
= 0.54 kg.m^{2}
(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90
= 0.026 x 0.6 x sin 90 = 0.0156 N
