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Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
The average force applied to the ball= 106.7 N
Explanation:
Force is given by
f= ΔP/t
ΔP= change in momentum= m Vf- m Vi
m= mass =0.2 kg
Vf= final velocity= 12 m/s
Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)
t= time= 60 ms= 0.06 s
now ΔP= 0.2 [ 12-(-20)]
ΔP=0.2 (32)=6.4 kg m/s
now force F= ΔP/t
F= 6.4/0.06
F=106.7 N
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Answer:
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