Answer:
D
Explanation:
<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>
Options A, B, and C requires that the forces of friction is increased in order to have more control.
However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.
<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>
The answer is 1,600 J.
A work (W) can be expressed as a product of a force (F) and a
distance (d):
W = F · d<span>
We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
Explanation:
Gravitational potential energy
= mgh
= (2kg)(10N/kg)(5m)
= 100J.
Answer:
31m/s
23.17°
Explanation:
Given the following :
From the diagram attached :
AB = 15m/s, BC = 18m/s, AC = a = resultant
Resolving Velocity into both vertical and horizontal component.
Kindly see attached picture for detailed explanation.
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
