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
Answer:
if I aint wrong it would 2nd one
Answer:
= 9.8°
Explanation:
Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.
width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m
angular position of fringe, Sinθ = n λ /a
n is order of fringe , λ is wave length of light and a is slit aperture
So Sinθ ∝ 1 / a
Sin θ₁ /Sin θ₂ = a₂/a₁ ;
Sin20°/sinθ₂ = 2 / 1
sinθ₂ = Sin 20° / 2 = .342/2 = .171
θ₂ = 9.8 °
Answer:
48 m
Explanation:
Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.
The eastbound train
First convert km/h to m/s
(97 × 1000)/3600
97000/3600
26.944444 m/s
As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion
V^2 = U^2 - 2as
O = 26.944^2 - 2 × 3.5 S
726 = 7S
S = 726/7
S1 = 103.7 m
The westbound train
Convert km/h to m/s
(127×1000)/3600
127000/3600
35.2778 m/s
Using third equation of motion
V^2 = U^2 - 2as
0 = 35.2778^2 - 2 × 4.2 × S
1244.52 = 8.4S
S = 1244.52/8.4
S2 = 148.2 m
S1 + S2 = 103.7 + 148.2 = 251.86
The distance between them once they stop will be
300 - 251.86 = 48.14 m
Therefore, the distance between them once they stop is 48 metres approximately.
