Work = Force * distance
Force = 70 N
Work = 3500 J
3500 = 70d
d = 3500/70 = 50 m
Answer:
v = 23.66 m/s
Explanation:
recall that one of the equations of motion may be expressed:
v² = u² + 2as,
Where
v = final velocity (we are asked to find this)
u = initial velocity = 0 m/s since we are told that it starts from rest
a = acceleration = 0.56m/s²
s = distance traveled = given as 500m
Simply substitute the known values into the equation:
v² = u² + 2as
v² = 0 + 2(0.56)(500)
v² = 560
v = √560
v = 23.66 m/s
Answer:
Option c. (Both Technician A and B are correct)
Explanation:
A transmission system consists of 3 shafts. The input shaft, the counter shaft, and the main shaft. The clutch gear always rotates with input shaft and is a crucial element of the input shaft.
The counter shaft is actually several gears machined out of a single piece of steel. The counter shaft may also be called counter gear or cluster gear. It is a secondary shaft that runs parallel to the mainshaft in a gearbox and is used to provide powers to machine components such as the drive axle.
The main gears (also called the speed gears) on main shaft (also known as the output shaft) are used to transfer rotation from counter shaft to the output shaft.
Hence in the light of above description, both technician A and B are correct.
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
I would say a. Good luck!
