Answer:
W = 3/2 n (T₁- T₂)
Explanation:
Let's use the first law of thermodynamics
ΔE = Q + W
in this case the cylinder is insulated, so there is no heat transfer
ΔE = W
internal energy can be related to the change in temperature
ΔE = 3/2 n K ΔT
we substitute
3/2 n (T₂-T₁) = W
as the work is on the gas it is negative
W = 3/2 n (T₁- T₂)
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.
Answer:
centripital acceleration= v^2/r
r = v^2/a
r=31.8×31.8/29
r=34.8703m
Explanation:
the maximum acceleration is obtained with minimum radius.
were solving for v velocity of the ball after it has hit the bottle. a. momentum ->p=mv->ball + bottle momentum during hit = ball + bottle momentum after hit-> ball (.5*21)+ bottle (.2*0) (it's 0 because the the bottle is standing still) = ball after hit (.5*v)+bottle after hit (.2*30) -> 10.5+0=.5v+6 ->4.5=.5v->v=9m/s
b. if bottle was heavier the ball would be slower so final velocity would decrease
Explanation:
momentum = mass x velocity
initial momentum = 100 x 15 = 1500kgm/s
after momentum = 100 x 20 = 2000kgm/s
a =(v-u)/t
a = (20-15)/10
a = 5/10
a = 0.5m/s²
f = ma
f = 100 x 0.5
f = 50N