I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m
Answer:
1.2 rad/s
Explanation:
m1 = 15 g, m2 = 9 g, ω1 = 0.75 rad/s
Let the new angular speed is ω2 and the radius of the table be r.
The angular momentum is conserved when no external torque is applied.
I1 ω1 = I2 ω2
(m1 + m2)x r^2 x 0.75 = m1 x r^2 x ω2
(15 + 9) x 0.75 = 15 x ω2
ω2 = 1.2 rad/s
Answer:
The force when θ = 33° is 1.7625 times of the force when θ = 18°
Explanation:
The force on a moving charge through a magnetic field is given by
F = qvB sin θ
q = charge of the moving particle
v = Velocity of the moving charge
B = Magnetic field strength
θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge
Because qvB are all constant, we can call the expression K.
F = K sinθ
when θ = 18°,
F = K sin 18° = 0.309K
when θ = 33°, let the force be F₁
F₁ = K sin 33° = 0.5446K
(F₁/F) = (0.5446K/0.309K) = 1.7625
F₁ = 1.7625 F
Hope this Helps!!!
Answer: C and D
Explanation: I’m pretty sure it’s C and D I got the same question but cold water let me know please, thanks!
