The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Drill Bit? This problem has been solved! See the answer. The high speed industrial drill with diameter of 98 cm develops 5.85hp at 1900 Rpm. What torque and force is applied to the drill bit?

1) Torque = 21.934 Nm

2) force = 44.76 N

Explanation:

Power = 5.88 hp

1 hp = 746 W

Power = 5.88 x 746 = 4364.1 W

Angular speed in Rpm = 1900 rpm

But angular speed w = (2¶N)/60 rad/s

= (2 x 3.142 x 1900)/60 = 198.968 rad/s

From,

1) Power P = T x w

Where T = torque

T = P/W = 4364.1/198.968 = 21.934 Nm

2) diameter of drill = 98 cm

Radius = 98/2 = 49 cm = 49x10^-2 m

From torque T = Force x radius

Force = Torque /radius

F = 21.934/49x10^-2 = 44.76 N

firearm identification is the process of analyzing the bullets and cartridge cases left at a crime scene to determine if they came from a particular firearm. marks on the bullets and cartridges may be common to every firearm of that type. these are called class characteristics.

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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