Answer:
B or D but im pretty sure it is D
Explanation:
When molecules are left in the sun, it heats up. When molecules heat up, the begin to vibrate rapidly. The sun is not constant as it could get blocked by clouds, so it would, at times, slow down the movement of the molecules. The answer is most likely D.
Answer:
Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.
.
Explanation:
Consider four possible cases.
<h3>Case A: 12.0 V.</h3>

In case all three capacitors are connected in parallel, the
capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.
<h3>Case B: 5.54 V.</h3>
![-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-](https://tex.z-dn.net/?f=-3.0%5C%3B%5Cmu%5Ctext%7BF%7D-%5B%5Cbegin%7Barray%7D%7Bc%7D-%7B%5Cbf%202.0%5C%3B%5Cmu%5Ctext%7BF%7D%7D-%5C%5C-1.5%5C%3B%5Cmu%5Ctext%7BF%7D-%5Cend%7Barray%7D%5D-)
In case the
capacitor is connected in parallel with the
capacitor, and the two capacitors in parallel is connected to the
capacitor in series.
The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.
The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,
.
What will be the voltage across the 2.0 μF capacitor?
The charge stored in two capacitors in series is the same as the charge in each capacitor.
.
Voltage is the same across two capacitors in parallel.As a result,
.
<h3>Case C: 2.76 V.</h3>
.
Similarly,
- the effective capacitance of the two capacitors in parallel is 5.0 μF;
- the effective capacitance of the three capacitors, combined:
.
Charge stored:
.
Voltage:
.
<h3 /><h3>Case D: 4.00 V</h3>
.
Connect all three capacitors in series.
.
For each of the three capacitors:
.
For the
capacitor:
.
Options:
A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.
B. Star 1 is 100 times more distant than Star 2.
C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.
D. Star 1 is 10 times more distant than Star 2.
E. Star 1 is 100 times nearer than Star 2.
Answer:
D. Star 1 is 10 times more distant than star 2
Explanation:
For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.
Luminosity, L = 1/r²
Where r is the distance of the star to the earth
Since star 1 is dimmer in brightness than star 2 by a factor of 100,
L₁/L₂ = 1/100
i.e. L₁ = 1, L₂=100
L₁ = 1/r₁² ............(1)
1 = 1/r₁²
L₂ = 1/r₂²
100 = 1/r₂² .........(2)
divide equation (2) by equation (1)
100/1 = ( 1/r₂² )/ (1/r₁²)
100 = (r₁/r₂)²
r₁/r₂ = √100
r₁/r₂ = 10
r₁ = 10r₂
<span>The unknown substance is silver.
I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so
273 g / 26 mL = 10.5 g/mL
Looking up a list of elements sorted by density, I see the following:
10.07 Actinium
10.22 Molybdenum
10.5 Silver
11.35 Lead
And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>
Answer:

Explanation:
According to the exercise we know the angle which the bait was released and its maximum height

To find the initial y-component of velocity we need to do the following steps:

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0


Since the bait is released at 25º

