Answer: Yes.
Explanation:
There is gravity in water because How does the water go up in the cycle? the garavity with the wind pull it up.
Answer:
11.87
Explanation:
final velocity^2= initial velocity ^2+ 2* Acceleration* distance
Final Velocity^2= 9*9+2*1.5.20
final velocity^2 = 141
final velocity = 11.87
Answer:
r = 255.68 m
Explanation:
When a body moves in a circular path, an acceleration, due to constant change in its direction, is developed, known as centripetal acceleration. The centripetal acceleration acts towards the center of the circular path. The formula to calculate the centripetal acceleration is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = 22 m/s²
v = tangential speed = 75 m/s
r = radius of curve = ?
Therefore,
22 m/s² = (75 m/s)²/r
r = (75 m/s)²/(22 m/s²)
<u>r = 255.68 m</u>
Answer:
equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )
Explanation:
Given data
mass = 3 slugs = 3 * 32.14 = 96.52 lbs
constant k = 9 lbs/ft
Beta = 6lbs * s/ft
mass is pulled = 1 ft below
to find out
equation of motion for the mass
solution
we know that The mass is pulled 1 ft below so
we will apply here differential equation of free motion i.e
dx²/dt² + 2 α dx/dt + ω² x =0 ........................1
here 2 α = Beta / mass
so 2 α = 6 / 96.52
α = 0.031
α² = 0.000961 ...............2
and
ω² = k/mass
ω² = 9 /96.52
ω² = 0.093 ..................3
we can say that from equation 2 and 3 that α² - ω² = -0.092239
this is less than zero
so differential equation is
x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )
equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )
Answer:
a) K_e = 0.1225 J, b) U = 1.96 J, c) v = 0.99 m / s
Explanation:
Let's use the simple harmonium movement expression
y = A cos (wt + Ф)
indicate that the amplitude is
A = 0.05 m
as the system is released, the velocity at the initial point is zero
v = dy / dt
v = - A w sin (wt + Ф)
for t = 0 s and v = 0 m/s
0 = - A w sin Ф
so Ф = 0
the expression of the movement is
y = 0.05 cos wt
The total energy of the system is
Em = ½ k A²
let's use conservation of energy
starting point. Spring if we stretch and we set the zero of our system at this point
Em₀ = K_e + U
Em₀ = 0
final point. When weight and elastic force are in balance
Em_f = K_e + U
Em_f = ½ k y² + m g (-y)
energy is conserved
Em₀ = Em_f
0 = ½ k y² + m g (-y)
k = 2mg / y
k = 2 4.00 9.8 / 0.050
k = 98 N / m
a) maximum elastic energy
K_e = ½ k A²
K_e = ½ 98 0.05²
K_e = 0.1225 J
b) the maximum gravitational energy
U = m g y
U = 4.00 9.8 0.05
U = 1.96 J
c) The maximum kinetic energy occurs when the spring is not stretched
U = K
mg h = ½ m v²
v = √2gh
v = √( 2 9.8 0.05)
v = 0.99 m / s
d) energy at any point
Em = K + U
