1.5 x 10^3 standard notation
1 answer:
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Answer:
t = 0.2845Nm (rounded to 4 decimal places)
Explanation:
The disk rotates at a distance of an arc length of 28cm
Arc length = radius × central angle × π/180
28cm = 10cm × central angle × π/180
Central angle = × 180/π ≈ 160.4°
Torque (t) = rFsin(central angle) , where F is the applied force
Radius in meters = 10/100 = 0.1m
t = 0.1m × 16N × sin160.4°
t = 0.2845Nm (rounded to 4 decimal places)
Answer:
a) Frope= 71.7 N
b) Frope=6.7 N
Explanation:
In the figure the skier is simulated as an object, "a box".
a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:
∑F=0
Frope=w*sen6.8°
Frope=71.71N
Take into account that w is the weight that is calculated as mass per gravitiy constant:
w=m*g
b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:
F=m*a
F=61.8Kg*0.109m/s2
Frope=6.73N
