Answer:
Force of Rope = 122.5 N
Force of Rope = 480.2N
Explanation:
given data
length = 3.00 m
mass = 25.0 kg
clown mass = 79.0 kg
angle = 30°
solution
we get here Force of Rope on with and without Clown that is
case (1) Without Clown
pivot would be on the concrete pillar so Force of Rope will be
Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 122.5 N
and
case (2) With Clown
so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be
Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 480.2N
Acceleration is the change in velocity over time
Answer:
D = 271.54 m
Explanation:
given,
1. car accelerates at 4.6 m/s² for 6.2 s
2. constant speed for 2.1 s
3. slows down at 3.3 m/s²
distance travel for case 1
using equation of motion
d₁ = 88.41 m
case 2
constant speed for 2.1 s now, we have to find velocity
v = u + at
v = 0 + 4.6 x 6.2
v = 28.52 m/s
distance travel in case 2
d₂ = v x t
d₂ = 28.52 x 2.1 = 59.89 m
for case 3
distance travel by the car
v² = u² + 2 a s
final velocity if the car is zero
0² = 28.52² + 2 x (-3.3) x d₃
6.6 d₃ = 813.39
d₃ = 123.24 m
total distance travel by the car
D = d₁ + d₂ + d₃
D = 88.41 + 59.89 + 123.24
D = 271.54 m
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
