Answer: 47.6 m/s
Explanation: Please see attached for the calculation and formula.
<u>The two ways to find acceleration in non uniform motion are as follows:</u>
<u>Explanation:</u>
Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).
Accordingly, non-uniform acceleration motion can be carried out in 2 ways:
Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.
The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.
Answer: when an ionic compound dissolves in water, the positive ends of the water molecules are attracted to the anions and the negative ends are attracted to the cations.
Hydrogen Chloride (HCl) is a compound which forms an acidic solution when dissolved in water.
Hope it helps :)
Explanation:
Answer:

Explanation:
Given that
V= 12 V
K=3
d= 2 mm
Area=5.00 $ 10#3 m2
Assume that
$ = Multiple sign
# = Negative sign

We Capacitance given as
For air







Net capacitance
C=C₁+C₂

We know that charge Q given as
Q= C V


I attached a free body diagram for a better understanding of this problem.
We start making summation of Moments in A,



Then we make a summation of Forces in Y,



At the end we calculate the angle with the sin.

