Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J
Answer:
Explanation:
Time taken by stone to cover horizontal distance
where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81
= 0.654654 seconds
t=0.65 s
Velocity, v= distance/time
v=10/0.65= 15.27525 m/s
v=15.3 m/s
where r is radius of circle, substituting r with 1.1m
Therefore, centripetal acceleration is
Answer:
2.19 N/m
Explanation:
A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:
T = 2π√(m/K)
Where T is the period, m is the mass (in kg), and K is the damping constant. So:
2.4 = 2π√(0.320/K)
√(0.320/K) = 2.4/2π
√(0.320/K) = 0.38197
(√(0.320/K))² = (0.38197)²
0.320/K = 0.1459
K = 2.19 N/m
Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.
Answer:(a)360N,(b)171N,(c)2.702m
Explanation:
(a)Maximum Friction Force =
=360 N
(b)Moment about Ground Point
(c)
Here maximum friction force can be 360 N
Therefore
Where x is the maximum distance moved by man along the ladder
740x=2000
x=2.702m