All the stars, spheres and galaxies that can be perceived nowadays
make up just 5 percent of the universe.
The former 95 percent is prepared of stuff stargazers can't see, notice or even
understand. These secretive substances are called dark energy and
dark matter. Experts determine their presence grounded on their gravitational
influence on what little bits of the universe can be perceived.
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
Answer:
2nd no. positive.... 3rd no. negitive......
Answer:
Ea = 112500[J]
Eb = 87500[J]
Explanation:
To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.
This can be easily visualized by the following equation:
Now we must define the energies at points A & B.
<u>For point A</u>
At point A we only have kinetic energy since it moves at 15 [m/s]
So the kinetic energy
![E_{A}=\frac{1}{2}*m*v_{A}^{2} \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]](https://tex.z-dn.net/?f=E_%7BA%7D%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7BA%7D%5E%7B2%7D%20%20%5C%5CE_%7BA%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2A1000%2A%2815%29%5E%7B2%7D%20%5C%5CE_%7BA%7D%3D112500%5BJ%5D)
The final kinetic energy can be calculated as follows:
![112500-25000=E_{B}\\E_{B}=87500[J]](https://tex.z-dn.net/?f=112500-25000%3DE_%7BB%7D%5C%5CE_%7BB%7D%3D87500%5BJ%5D)
The time taken by the ballast bag to reach the ground is 2.18 s
The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity <em>u </em>and it falls under the acceleration due to gravity<em> g.</em>
Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.
The bag makes a net displacement <em>s</em> of 13.4 m downwards, hence
Its initial velocity is
The acceleration due to gravity acts downwards and hence it is negative.
Use the values in the equation of motion and write an equation for t.
Solving the equation for t and taking only the positive value for t,
t=2.18 s
