Answer: The speed of the edge of the wheel = 19.72m/s
Explanation:
The liner speed of the edge of the wheel is given as;
V=wr
V= linear speed =?
w= angular velocity =456rpm
Converting rpm to rad/s
1rpm = π/30 rad/s
456rpm = 47.75 rad/s
r=0.413m
V=47.75 × 0.413
V= 19.72m/s
Therefore speed of the edge of the wheel = 19.72m/s
The force required to pull the two hemispheres is 46622.72N
<h3>Calculation and Parameters</h3>
( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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Answer:
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hope it helps !
That is called the capacity.
