I don’t know what to do

I Dont Know What To Do

Answers

Answer 1

Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.


Related Questions

You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the mean value of your results to three significant digits. ________

Answers

Answer: The mean value = 9.85m/s².

Explanation:

Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]

The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.

Number of measurements =9

Sum of measurements =  88.69

Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]

Hence, the mean value = 9.85m/s².

What is unusual about the material that Emily invented?

Answers

Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

Hope it helps! Please mark brainliest.

Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

a 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
how much work did it take for the student to travel from the ground to the top of the platform diving board? ​

Answers

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

Agnes makes a round trip at a constant speed to a star that is 16 light-years distant from Earth, while twin brother Bert remains on Earth. When Agnes returns to Earth, she reports that she has celebrated 20 birthdays during her journey. (a) What was her speed during her journey

Answers

Answer:

Speed of Agnes during her journey was 0.848c

Explanation:

Given that;

Age of Agnes t₀ = 20 years

distance d = 2 × distance of star from Earth = 2 × 16 light-years

= 32 light-years

so get her speed speed; we use the following expression

Yvt₀ = d

( v / √(1 - ([tex]\frac{v}{c}[/tex])²) )² = ( 32 light-years / 20 yrs )²

v² / (1 - ( v²/c²)) = ( 32 × c  / 20 )²

v² / (1 - ( v²/c²)) = 2.56 × c²  

v² / c²-v²/c² = 2.56 × c²  

v²c²  / c² - v²  = 2.56 × c²  

v²  / c² - v²  = 2.56

v²  = 2.56 (c² - v²)

v²  = 2.56 (c² - v²)

v²  = 2.56c² - 2.56v²

v² + 2.56v² = 2.56c²

3.56v² = 2.56c²

v² = (2.56/3.56)c²

v = √((2.56/3.56)c²)

so v = 0.848c

Therefore, Speed of Agnes during her journey was 0.848c

A 10 kg remote control plane is flying at a height of 111 m. How much
potential energy does it have?

Answers

Answer:

10.88kJ

Explanation:

Given data

mass= 10kg

heigth= 111m

Applying

PE= mgh

assume g= 9.81m/s^2

substitute

PE= 10*9.81*111

PE=10889.1 Joules

PE=10.881kJ

Hence the potential energy is 10.88kJ

Current Attempt in Progress The atomic radii of a divalent cation and a monovalent anion are 0.52 nm and 0.125 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance

Answers

Answer:

a)   F = 1.70 10⁻⁹N,   F = 1.47 10⁻⁸ N,

b) * the electronegative repulsion, from the repulsion by quantum effects

Explanation:

a) The atraicione force comes from the electric force given by Coulomb's law,

           F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]

divalent atoms

In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given

           F = k q² / r²

           F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]

           F = 1.70 10⁻⁹N

monovalent atoms

in this case the load is q = q₀

           F = 2 \ 10^9 \  \frac{ (1.6 \  10^{-19} )^2}{ (0.125 10^{-9} )^2 }

           F = 1.47 10⁻⁸ N

b) repulsive forces come from various sources

* the electronegative repulsion of positive nuclei

* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together

* from the repulsion of electron clouds, by quantum effects

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