Un alambre de 2m de longitud soporta una carga de 5400 N. El diámetro de la sección transversal es de 2 mm y el módulo de Young
1 answer:
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Centripetal Force (Fcp) = ?
His arm length = Radius (R) = 0.75 m
Discus velocity = Linear Velocity (V) = 5 m/s
Discus mass (m) = 2 kg
Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).
Answer: Last option, 66 N.
His arm length = Radius (R) = 0.75 m
Discus velocity = Linear Velocity (V) = 5 m/s
Discus mass (m) = 2 kg
Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).
Answer: Last option, 66 N.
Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill,
We need to find the y component of the normal force. We know that the y component of the normal force is given by :
So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.