What is the kinetic energy of a 2000 kg car traveling at a speed of 30 m/s (?65 mph)?
1 answer:
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For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
<h3>What is the slope of end a of the cantilevered beam?</h3>Generally, the equation for the is mathematically given as
Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
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Answer:
The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
Explanation:
The centripetal acceleration is given by:
Where:
: is the tangential speed = 9.50 m/s
r: is the distance = 6.00 m
Hence, the centripetal acceleration is:
Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
I hope it helps you!