A)After 3,5s -->v=v0+gt=21+(-9,8•3,5)=21+(-34,3)=-13,3m/s;
b)The maximum height that the orange reaches is h max=v0^2/2g=22,5m;v^2=sqrt(2gh)=>h=v^2/2g=9,025m.The height of the orange is H=h max-h=13,475m.;
c)The orange is traveling down.
Answer:
23.98 rpm
Explanation:
d = diameter of merry-go-round = 2.4 m
r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m
m = mass of merry-go-round = 270 kg
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²
M = mass of john = 34 kg
Moment of inertia of merry-go-round and john together after jump is given as
I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²
w = final angular speed
w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s
v = speed of john before jump
using conservation of angular momentum
Mvr + I w₀ = I' w
(34) (5) (1.2) + (194.4) (2.093) = (243.36) w
w = 2.51 rad/s
w = 23.98 rpm
Answer: C
produces as much as or more energy than it uses
Explanation:
This implies that the total energy used by the building is equivalent to the energy generated by the site
Explanation:
Below is an attachment containing the solution
Answer:
Radius of the outer most dark fringe is 2.65 cm
Solution:
As per the question:
Radius of curvature of the glass, r = 10.8 m
No. of dark fringes, n = 100
Wavelength of light,
Now,
To calculate the radius R of the outermost ring:
Radius of the dark fringe of nth order is given by: