A squirrel sitting on the ground starts to run with an acceleration of 5.1 m/s 2. How fast is the squirrel going after 0.25 seco
2 answers:
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Answer:
f = 692 N
Explanation:
given data:
f =800N
a =1.2 m s^{2}
m= 90 kg
from newton's second law
net force
therefore we have from above equation
ma =F - f
putting all value to get force of friction
1.2*90 = 800 - f
f = 692 N
Answer:
If a vertical line extending down from an object's CG extends outside its area of support, the object will topple
Explanation:
We can understand better this situation using a diagram with the forces acting on it.
In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.
