Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ =
τ =
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω =
ω =
ω =
A = 0.52 rad/s B = -0.182 rad/s²
Answer:
Maximum height of rocket = 2538.74 m
Explanation:
We have equation of motion s = ut + 0.5 at²
For first 5 seconds
s = 0 x 5 + 0.5 x 40 x 5² = 500 m
Now let us find out time after 5 seconds rocket move upward.
We have the equation of motion v = u + at
After 5 seconds velocity of rocket
v = 0 + 40 x 5 = 200 m/s
After 5 seconds the velocity reduces 9.8m/s per second due to gravity.
Time of flying after 5 seconds
Distance traveled in this 20.38 s
s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m
Maximum height of rocket = 500 +2038.74 = 2538.74 m
Answer:
The resultant displacement of plane = 1310.76 Km
Explanation:
Given that,
Speed of the plane = 925 Km/h
Therefore, in 1 hour plane travels 925 Km
From figure,
Displacement if it travels east for 1.5 h, AB = 1387.5 Km
Displacement if it travels north-west for 2 h, BC = 1850 Km
Angle formed between two displacement = 135°
The resultant displacement is given by the relation
AC² = AB² + BC² + 2 AB BC Cos θ
Substituting the values in the equations,
AC² = 1387.5² + 1850² + 2 x 1387.5 x 1850 x Cos 135°
= 1718095
AC = 1310.76
Therefore, the resultant displacement of plane is 1310.76 Km
