M= 1000 g g= 10 m/s2 h= 10 K.E at GROUND = 400 j M.E = ....... ........J the answer is 400 so plz explain how
1 answer:
You might be interested in
Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :
So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.
Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = ( × mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s
Answer:
1.3 x 10⁻⁴ m
Explanation:
= wavelength of the light = 450 nm = 450 x 10⁻⁹ m
n = order of the bright fringe = 1
θ = angle = 0.2°
d = separation between the slits
For bright fringe, Using the equation
d Sinθ = n
Inserting the values
d Sin0.2° = (1) (450 x 10⁻⁹)
d (0.003491) = (450 x 10⁻⁹)
d = 1.3 x 10⁻⁴ m