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mestny [16]
3 years ago
9

While flying over the grand canyon the pilot

Physics
1 answer:
Veronika [31]3 years ago
6 0
We cannot answer the question if we do not have the answers - sorry!!
You might be interested in
Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the
Brrunno [24]

Answer:

13.83928\times 10^{-6}\ F

249.10704\times 10^{-6}\ J

137.89848\times 10^{-6}\ J

111.20842\times 10^{-6}\ J

2.98273 V

Explanation:

C_1=25\ mu F

C_2=31\ mu F

V = Voltage = 6 V

Equivalent capacitance is given by

\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F

Equivalent capacitance is 13.83928\times 10^{-6}\ F

Energy stored is given by

E=\dfrac{1}{2}CV^2\\\Rightarrow E=\dfrac{1}{2}\times 13.83928\times 10^{-6}\times 6^2\\\Rightarrow E=249.10704\times 10^{-6}\ J

Total energy stored is 249.10704\times 10^{-6}\ J

Charge is given by

Q=CV\\\Rightarrow Q=13.83928\times 10^{-6}\times 6\\\Rightarrow Q=83.03568\times 10^{-6}\ C

Voltage is given by

V_1=\dfrac{Q}{C_1}\\\Rightarrow V_1=\dfrac{83.03568\times 10^{-6}}{25\times 10^{-6}}\\\Rightarrow V_1=3.3214272\ V

E_1=\dfrac{1}{2}C_1V_1^2\\\Rightarrow E_1=\dfrac{1}{2}\times 25\times 10^{-6}\times 3.3214272^2\\\Rightarrow E_1=137.89848\times 10^{-6}\ J

Energy strored in C1 is 137.89848\times 10^{-6}\ J

V_2=\dfrac{Q}{C_2}\\\Rightarrow V_2=\dfrac{83.03568\times 10^{-6}}{31\times 10^{-6}}\\\Rightarrow V_2=2.67857\ V

E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J

Energy stored in C2 is 111.20842\times 10^{-6}\ J

E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J

So, the energy is equivalent

Equivalent capacitance

C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F

E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V

The voltage would be 2.98273 V

8 0
4 years ago
Two small, positively charged spheres have a combined charge of 5.23 x 10^-5 C. If each sphere is repelled from the other by an
pshichka [43]

Answer:

The smallest charge of the dial is 2.46 10-5 C

Explanation:

The Coulomb force is responsible for the electroactive repulsion, the equation that describes it is

       

          F = k q1 q2 / r²

Where K is the Coulomb constant that is worth 8.99109 N m² / C², q is the electric charge of each sphere and r is the distance between them.

They also give us the condition that the sum of the charge is 5.23 10-5 C

          Qt = q1 + q2 = 5.23 10⁻⁵ C

Let's replace in the Coulomb equation, let's clear and calculate

       

         F = k (Qt -q2) q2 / r²

         F = k q22 / r² - k Qt q2 / r²

         1.0 = 8.99 10⁹ q2² /2.04² - 8.99 10⁹ 5.23 10⁻⁵ q2 / 2.04²

         1.0 = 2.16 10⁹ q2²2 - 11.30 10⁴ q2

         0 = 2.16 109 q2² - 11.30 10⁴ q2 -1.0                (* 1/2.16 109)

          0 = q2² - 1.05 10⁻⁵ q2 - 0.463 10⁻⁹

Let's solve the second degree equation for q2

         q2 = 1.05 10⁻⁵ ±√[(1.05 10⁻⁵)² - 4 1 (-0.463 10⁻⁹)] / 2

         q2 = 1.05 10⁻⁵ ±√ [1.10 10⁻¹⁰ + 18.52 10⁻¹⁰] / 2

         q2 = {1.05 10⁻⁵ ± 4.43 10⁻⁵} / 2

The solutions are

        q2 ’= 2.74 10-5 C

        q2 ’’ = -1.69 10-5 C

As the problem tells us that the spheres are positively charged, the correct solution is 2.74 10-5 C, let's see the charge of the other sphere

                  Qt = q1 + q2 ’

                  q1 = Qt -q2 ’

                  q1 = 5.23 10-5 - 2.74 10-5

                  q1 = 2.46 10-5 C

The smallest charge of the dial is 2.46 10-5 C

5 0
4 years ago
What forms of energy does a pendulum utilize?
jonny [76]
It holds kinetic and potential energy
8 0
4 years ago
A spring that has a spring constant of 440 N/m exerts a force of 88 N on a box. What is the displacement of the spring? 0. 2 m 5
mina [271]

The displacement of a spring is directly proportional to the applied force. The displacement of the given spring is 50.45 m.

<h2></h2><h2>From Hooke's law:</h2>

F = kx

Where,

F - force applied = 88 N

k - spring constant = 440 N/m

x - displacement = ?

Put the values in the formula,

88  = 440 \times x\\\\x = \dfrac {440 }{88}\\\\x = 50.45 \rm \ m

Therefore, the displacement of the given spring is 50.45 m.

Learn more about Hooke's law:

brainly.com/question/3355345

6 0
3 years ago
Which is an example of qualitative data?
LiRa [457]
The answer is C)the rating that the golfers give Callaway clubs
6 0
3 years ago
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