Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data




Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)

The increase in potential energy of the ball is 115.82 J
Answer: The diameter of the circular path is 2.96m
Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.
Centripetal acceleration = 2.7m/s^2
Tangential speed = 2.0m/s
Radius = 2.0^2 / 2.7 = 4/2.7
= 1.48m
Diameter = radius*2
= 1.48*2 = 2.96m.

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
Answer:
Dx = -0.5
Dy = -0.25
Explanation:
Two vectors are given in rectangular components form as follows:
A = i + 6j
B = 3i - 7j
It is also given that:
A - B - 4D = 0
so, we solve this to find D vector:
(i + 6j) - (3i - 7j) - 4D = 0
- 2i - j = 4D
D = - (2/4)i - (1/4)j
D = - (1/2)i - (1/4)j
<u>D = - 0.5i - 0.25j</u>
Therefore,
<u>Dx = -0.5</u>
<u>Dy = -0.25</u>