Answer:
1) U-> K +W
2) K -> W
Explanation:
In this exercise, care must be taken as they indicate that the friction force (rubbing) is not negligible.
1 part at the top of the hill the car has gravitational potential energy, which is transformed in a part into kinetic energy and another part into heat by the work of the friction force that opposes the movement.
2 part when the other hill rises it loses kinetic energy that is transformed into gravitational potential energy and part in heat due to the work of the friction force on this hill.
3rd part in the last descent all the gravitational potential energy is transformed into kinetic energy and the application of the brakes is transformed into heat due to the negative lock of the friction force and filled with gasoline that has chemical energy that can be transformed in the engine.
Answer:
The fingertips should be at point of attachment of the cuff to the sleeve
Explanation:
In the closed cuff method, the gown is picked up from the wrapper by holding onto the exposed inside top layer.
The picked up gown should be handled by holding the region close to the neck of the gown, at the same time avoiding contact of the gown with your body or other objects that are unsterile
The arms are then slid into the gown sleeves with the hands at the shoulder level to avoid contact with the body of the gown
The arms are then further slid into the sleeves with the assistance of the circulator to the point where your fingertips are at the midpoint of the attachment of cuff and sleeve
<span>So we want to know how many satellites exist in Earth's orbit today. If by satellites we mean artificial vehicles made to orbit the Earth, then, according to Goddard Space Flight Center's list, there is 2271 satellite in the Earths orbit today. So the correct answer is 2271. </span>
Answer:
F_total = 29.4 N, directed to the right of particle 2
Explanation:
We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.
Let's use Coulomb's law to calculate each force
F = 
particles 1 and 2
q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m
F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²
F₁₂ = 2.59 10¹ N
Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.
particles 2 and 3
q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m
we calculate
F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²
F₂₃ = 3.5 N
as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2
Now we add the forces as vectors
F_total = ∑ F = F₁₂ + F₂₃
F_total = 25.2 +3.5
F_total = 29.4 N
directed to the right of particle 2
