By using Kepler's 3rd law we find that A year on Earth is shorter than a year on Saturn (d)
<h3>Further explanation</h3>Kepler's law is based on the orbits of planets around the sun.
There are 3 laws issued by Kepler
the inner planet's orbit around the sun is based on an elliptical path and the sun is the center
Imaginary lines are drawn from planets to the sun cover the same area at the same time interval.
The square of the orbital period of a planet is proportional to the cube of its average distance to the sun
General formula:
There are choices of questions that need to be completed from the questions above:
a. A day on Earth is shorter than a day on Saturn.
b. A year on Earth is longer than a year on Saturn.
c. A day on Earth is longer than a day on Saturn.
d. A year on Earth is shorter than a year on Saturn.
If we use Kepler 3's Law to see the answer choices then:
Kepler 3's law is based only on the period of the planet, in this case, the planet's revolution against the sun.
So the determination of the length of the day can not be determined using Kepler 3 Law
So answers a and c do not meet the area of Kepler 3 law
One year is counted one revolution = one cycle = one orbit
The longer the planet's distance from the sun, the longer it will get one rotation so that the longer the year from the planet
The planets that circle the sun in sequence from the closest are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune
From this sequence shows that the distance of the planet Saturn is further than that of the planet Earth so
A year on Earth is shorter than a year on Saturn (d)
<h3>Learn more</h3>The Sun has more gravity than the other plans in our solar system
weight and mass used synonymously on Earth
the Earth has more gravitation pull than the moon
Keywords: Kepler law 3, sun, revolution, earth, Saturn, rotation, year, day
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a =
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
Answer:
a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,
e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J
Explanation:
In a mass-spring system with simple harmonic motion, the angular velocity is
w =
a) find the period
angular velocity, frequency, and period are related
w = 2π f = 2π / T
f = 1 / T
T = 1 / f
T = 1 / 0.59
T = 1.69 s
b) the spring constant
w = 2π f
w = 2π 0.59
w = 3.70 rad / s
w² = k / m
k = w² m
k = 3.70² 0.060
k = 0.825 N / m
c) the maximum speed
simple harmonic movement is described by the expression
x = A cos (wt + Ф)
speed is defined by
v =
v = -A w sin (wt + fi)
the speed is maximum when the cosine is ± 1
v = A w
v = 0.394 3.70
v = 1.46 feet/s
d) maximum acceleration
a =
a = - A w² cos wt + fi
the acceleration is maximum when the cosine is ±1
a = A w²
a = 0.394 3.70²
a = 5.41 ft / s²
e) velocity and acceleration for x = 6 cm
let's reduce the cm to feet
x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot
Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s
let's use the expression for the velocity
v = -A w sin (0 + Фi)
0 = - A w sin Ф
so sin Ф = 0 which implies that Фi = 0
the equation of motion is
x = A cos wt
x = 0.394 cos 3.70t
we substitute
0.1969 = 0.394 cos 370t
3.70 t = cos⁻¹ (0.1969 / 0.394)
let's not forget that the angle is in radians
3.70, t = 1.047
t = 1.047 / 3.70
t = 0.2826 s
we substitute this time in the equation for velocity and acceleration
v = - Aw sin wt
v = - 0.394 3.70 sin 3.70 0.2826
v = - 1,319 ft / s
a = - A w² cos wt
a = - 0.394 3.70² cos 3.70 0.2826
a = - 2.70 ft / s²
f) the kinetic and potential energy at this point
K = ½ m v²
let's slow down to the SI system
v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s
K = ½ 0.060 0.402²
K = 4.8 10⁻³ J
U = ½ k x²
U = ½ 0.825 0.06²
U = 1.49 10⁻³ J