Answer:
1/√2
Explanation:
Kinetic energy of a body is expressed as
KE = 1/2mv²
If it's KE is doubled
2KE = 1/2mv1² where v1 is the new speed when the kinetic energy is doubled
To know the value of the amount the velocity has changed, we will divide both equations
2KE/KE = (1/2mv²)/(1/2mv1²)
2 = v²/v1²
(v/v1)² = 2
v/v1 = √2
v1 = v/√2
v1 = 1/√2 × v
The new velocity has changed by
1/√2vinitial
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
Answer:
96%
Explanation
Let A the total area of the galaxy, is modeled as a disc:
A = πR^2 = π (25 kpc)^2
And let a be the area that astronomers are able to see:
a = πr^2 = π(5 kpc)^2
The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:
P = 100 a/A = (5/25)^2 = 100/25 = 4%
Therefore, the percentage of the galaxy not included, i.e. not seen is:
(100-4)% = 96%
Answer: I would say the object with the Lower velocity because Lighter with Higher velocity makes it heavy, velocity is how heavy something is so the lighter it is the less difficult it will be to catch.
