Answer:
The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>
Explanation:
As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.
<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>
Where,
<em>R = Orbital radius (in this case) = 30.1 AU</em>
<em />
Plug the value of R in the equation (A):
<em>(A) => The circumference of the circle = 2*π*(30.1)</em>
<em> The circumference of the circle = </em><em>60.2π</em>
Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>
Lab safety equipment prevents damage from accidents and helps keep the people working in the lab safe. The equipment goes hand in hand with the clothing of the person. The first step would be to wear closed shoes and a lab coat.
The equipment that must be worn are goggles to protect the eyes from irritants and latex gloves to protect the skin on the hands.
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm =
- Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m