D. Heat energy will be transferred within the system and if left long enough, there will be enough transferred energy to make both of them the same temperature.
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula 
E = 1461.95 N/C
c) The electric field E is calculated as:
E = 239.76 N/C
A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
Therefore the correct answer is "all of the above".
b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
The correct answer is again "all of the above".
Is there a equation or something so I can do the math of how many flowers there are at the end of the two monthsm
Answer:
F₄ = 29.819 N
Explanation:
Given
F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N
F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N
F₃ = (0 i + 0 j + 4 k) N
Then we have
F₁ + F₂ + F₃ + F₄ = 0
⇒ F₄ = - (F₁ + F₂ + F₃)
⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N
The magnitude of the force will be
F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N
