Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
Answer:
Convection occurs when thermal energy is transferred by the movement of fluid particles.
Explanation:
Transfer of heat energy by the movement of fluid particles is called convection. Convection takes place in liquids and gases due to kinetic energy. When heat is provided to liquids and gases they expand and move faster.
Molecules with higher kinetic energy become less dense and rise up to the surface of liquid, whereas molecules with lower energy move towards the bottom. This up and down movement of molecules causes convection currents in fluids.
Answer:
The displacement of the air drop after 3 second is 18.27 m.
Explanation:
Mass of the rain drop = m = 
Weight of the rain drop = W
Duration of time = t = 3 seconds
Drag force on rain drop = 
Motion of the rain drop:
Net force on the rain drop , F= W - D
v = 12.18 m/s
Initial velocity of the rain drop = u = 0 (since, it is starting from rest)
v=u+at (First equation of motion)
(second equation of motion)
s = 18.27 m
The displacement of the air drop after 3 second is 18.27 m.
Answer:
t = 0.354 hours
Explanation:
given,
coefficient of rolling friction μr=0.002
mass of locomotive = 180,000 Kg
rolling speed = 25 m/s
The force of friction = μ mg
= (.002) x (180000) x (9.8)
= 3528 N
F = m a
now,
m a = 3528 N
180000 x a = 3528
a = 0.0196 m/s²
Then apply
v = u + at
0 = 25 - 0.0196 x t
t = 1275.51 sec
t = 1275.61/3600 hours
t = 0.354 hours
time taken by the locomotive to stop = t = 0.354 hours
