Asked by Morty
                A 2.0Kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in .50s. What net force acts on the otter along the incline?
            
            
        Answers
                    Answered by
            drwls
            
    The force (F) equals the mass (M= 2.0 kg) times the acceleration.
If it travels 0.85 m in t = 0.50 s, the acceleration a is given by solving
0.85 = (1/2) a t^2 = = (a/2)*0.25
a = 6.8 m/s^2
Solve for F
    
If it travels 0.85 m in t = 0.50 s, the acceleration a is given by solving
0.85 = (1/2) a t^2 = = (a/2)*0.25
a = 6.8 m/s^2
Solve for F
                    Answered by
            tyreek
            
    13.6 n
    
                    Answered by
            Hala
            
    givens:M=2.0kg,vi=0m/s,displacement=85cm(0.85m),t=0.50s.
formula=average acceleration=average velocity/total time
>>average velocity=displacement/time
>>average velocity=0.85/0.5=1.7m/s
>>average acceleration=1.7/0.5=3.4m/s^2
force=mass*acceleration
force=2.0*3.4=6.8N
    
formula=average acceleration=average velocity/total time
>>average velocity=displacement/time
>>average velocity=0.85/0.5=1.7m/s
>>average acceleration=1.7/0.5=3.4m/s^2
force=mass*acceleration
force=2.0*3.4=6.8N
                    Answered by
            Cyrus B. Vonyon
            
    6.8n
    
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