This question involves the concepts of equilibrium and Newton's third law of motion.
The support force will be "1 pound" for the empty bucket and the support force will be "6 pounds" after pouring water into it.
- According to the condition of equilibrium, the sum of forces acting on a stationary object must be zero. Hence, the support force of the table will be equal to the total mass of the bucket.
- According to Newton's Third Law of Motion every action force has an equal but opposite reaction force. Hence, the support force will be a reaction force to the weight of the bucket.
Therefore, the support force in each case will be equal to the total mass of the bucket:
Case 1 (empty bucket):
<u>support force = 1 pound</u>
<u></u>
Case 1 (water poured):
support force = 1 pound + 5 pound
<u>support force = 6 pound</u>
<u></u>
Learn more about equilibrium here:
brainly.com/question/9076091
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
Answer:
(A) No
(B) Speed decreases
Explanation:
(A) since there is nothing propelling the boat and the friction between the ice and the boat and also air resistance is negligible the net force of the system in the horizontal direction is zero and hence there is no change in the horizontal momentum of the boat.
(B) Since the person had not velocity in the horizontal direction before landing on the boat but now has one after landing on the boat, the speed of the boat will decrease because the momentum has to be conserved (remember there is no change in it).
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
