Answer:
Minimum speed will be equal to 2.213 m/sec
Explanation:
We have given radius of the r = 2 m
Coefficient of friction 
At minimum speed frictional force will be equal to centripetal force
So 
So the minimum speed will be equal to 2.213 m/sec
We're not completely sure what vEp stands for.
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Light, being both a wave and a particle, displays many properties similar to both of them. Light, like waves, travels at different speeds through mediums because they provide any variable amount of resistance. Even air resists light to a degree, and light speed through air is slower than light speed in a vacuum.
The stiffness constant of the spring is 68,290.3 N/m
<h3>
Stiffness constant of the spring</h3>
Apply the principle of conservation of energy;
U = K.E
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
where;
- v is speed = 60 km/h = 16.67 m/s
- x is the distance
k = (1300 x 16.67²)/(2.3²)
k = 68,290.3 N/m
Thus, the stiffness constant of the spring is 68,290.3 N/m.
Learn more about stiffness constant here: brainly.com/question/1685393
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The answer is D=M/V hope it helps!!
