Answer:
The track's angular velocity is W2 = 4.15 in rpm
Explanation:
Momentum angular can be find
I = m*r^2
P = I*W
So to use the conservation
P1 + P2 = 0
I1*W1 + I2*W2 = 0
Solve to w2 to find the angular velocity
0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2
W2 = 0.435 rad/s
W2 = 4.15 rpm
I believe the correct answer is atmosphere (D).
Answer:
force for start moving is 7.49 N
force for moving constant velocity 2.25 N
Explanation:
given data
mass = 7.65 kg
kinetic coefficient of friction = 0.030
static coefficient of friction = 0.10
solution
we get here first weight of block of ice that is
weight of block of ice = mass × g
weight of block of ice = 7.65 × 9.8 = 74.97 N
so here Ff = Fa
so for force for start moving is
Fa = weight × static coefficient of friction
Fa = 74.97 × 0.10
Fa = 7.49 N
and
force for moving constant velocity is
Fa = weight × kinetic coefficient of friction
Fa = 74.97 × 0.030
Fa = 2.25 N
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 °