
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
V=IR
Potential Difference (v)= Current (A) * Resistance (Ω)
As V increases, R also increases.
Answer:
1.5 m
Explanation:
Let the distance from the box to the pivot be c.
Let the distance from the pivot to the effort be y.
From the question given above, the following data were obtained:
Effort force (Fₑ) = 7 N
Force of resistance (Fᵣ) = 14 N
Distance from the box to the pivot (c) = 0.75 m
Distance from the pivot to the effort (y) =?
Clockwise moment = Fₑ × y
Anticlock wise moment = Fᵣ × c
Clockwise moment = Anticlock wise moment
Fₑ × y = Fᵣ × c
7 × y = 14 × 0.75
7 × y = 10.5
Divide both side by 7
y = 10.5 / 7
y = 1.5 m
Therefore, the distance from the pivot to the effort is 1.5 m
Answer:
33.516 kJ
Explanation:
Potential energy is given by:
PE = mgh
Where m is the mass, g is acceleration due to gravity, and h is the height. In this case:
PE = 38kg x 9.8m/s^2 x 90m = 33516 kg m^2/s^2 = 33516 J = 33.516 kJ