Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m
Answer:
2a.) Wavelength = 1.8 m
2b.) F = 66.67 Hz
3a.) Find the attached file
3b.) Wavelength = 0.6 m
Explanation:
Given that the
Length L = 0.9m
Wavelength (λ) = 2L/n
Where n = number of harmonic
If n = 1, then
Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m
b.)
If waves travel at a speed of 120m/s on this string, what is the frequency
associated with the longest wave (first harmonic)?
Given that V = 120 m/s
V = Fλ
But λ = 2L, therefore,
F = V/2L
F = 120/1.8
F = 66.67 Hz
3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic
Using the same formula,
Wavelength (λ) = 2L/n
Where n = 3
Wavelength (λ) = 2 × 0.9/3
Wavelength (λ) = 0.6 m
Answer:

Explanation:
We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

In this formula,
is the final velocity,
is the initial velocity,
is the acceleration, and
is the time.
The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

Substitute the values into the formula.

Solve inside the parentheses.

Add.

The units can also be written as:

The bicycle's final velocity is 15 meters per second.
Answer: 146.1 kg/m/s
Explanation: Momentum is equal to mass times velocity, so you do 9.74 kg times 15 m/s and get 146.1 kg/m/s.
Electromagnetic waves can travel through a vacuum. Because they do not give up energy
in traveling, Electromagnetic waves can cross the great distances that separate stars
and galaxies. Rays from the Sun travel nearly 150 million
kilometers to reach Earth's surface.