Answer:
12900 W
24200 W
Explanation:
Given:
v₀ = 0 m/s
v = 1.3 m/s
t = 2.0 s
Find: a and Δx
v = at + v₀
(1.3 m/s) = a (2.0 s) + (0 m/s)
a = 0.65 m/s²
Δx = ½ (v + v₀) t
Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)
Δx = 1.3 m
While accelerating:
Newton's second law:
∑F = ma
F − mg = ma
F = m (g + a)
F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)
F = 19855 N
Power = work / time
P = W / t
P = Fd / t
P = (19855 N) (1.3 m) / (2.0 s)
P ≈ 12900 W
At constant speed:
Newton's second law:
∑F = ma
F − mg = 0
F = mg
F = (1500 kg + 400 kg) (9.8 m/s²)
F = 18620 N
Power = work / time
P = W / t
P = Fd / t
P = Fv
P = (18620 N) (1.3 m/s)
P ≈ 24200 W
Answer:
m = image distance/object distance
m=35/15=7/3
f=uv/(u+v)
35*15/(35+15)
f=10.5cm
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:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s
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