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3241004551 [841]
2 years ago
5

Consider two capacitors with unequal capacitance connected in parallel to a battery. Which of the following statements are true?

Physics
1 answer:
Allisa [31]2 years ago
3 0

Answer:

  1. The equivalent capacitance of the combination is greater than the capacitance of either of the capacitors.
  2. The voltage across each of the capacitors is the same.
  3. The sum of the charge stored on each capacitor is equal to the charge supplied by the battery.

Explanation:

The capacitance connected in parallel will have the same potentials across its ends. If the battery has a charge Q, it is divided among the capacitors.

That is,

                      Q = q₁ + q₂ + q₃

But, the potential is shared equally as V. So, the individual capacitance of the equation has the form q₁ = C₁V , q₂ = C₂V, etc.

                   

So, in this case, when the effective capacitance is formulated, it would be

                          C = C₁ + C₂ + C₃     farad

Therefore, the true statements are

The equivalent capacitance of the combination is greater than the capacitance of either of the capacitors.

The potential across each of the capacitors is the same.

The charge supplied by the battery divided among the capacitors.

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If you take any pitch on the keyboard, the next occurrence of the same letter name going towards the left (down) will vibrate:
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Answer:

A. Twice as slow

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A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a
Korvikt [17]

Answer:

v^2=v_o^2-2\times a\times (v_o.t)

Explanation:

Given:

Initial velocity of the vehicle, v_o

distance between the car and the tree, x_f

time taken to respond to the situation, t

acceleration of the car after braking, a

Using equation of motion:

v^2=u^2+2a.s ..............(1)

where:

v= final velocity of the car when it hits the tree

u= initial velocity of the  car when the tree falls

a= acceleration after the brakes are applied

s= distance between the tree and the car after the brakes are applied.

s=v_o\times t

Now for this situation the eq. (1) becomes:

v^2=v_o^2-2\times a\times (v_o.t) (negative sign is for the deceleration after the brake is applied to the car.)

5 0
2 years ago
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
schepotkina [342]
If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!
5 0
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Under which one of the following circumstances will heat transfer occur via convection? Group of answer choices Convection occur
BartSMP [9]

Answer:

1) in metal object heat transfer through the conduction .In vacuum heat transfer through only radiation .

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8 0
2 years ago
Read 2 more answers
Calculate the critical angle between glass (n = 1.90) and ice (n = 1.31)? 43° 52° 60° 75°
Ket [755]

Answer: 43 \°

Explanation:

According to <u>Snell’s Law</u>:  

n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})

Where:  

n_{1}=1.90 is the first medium index of refraction  (glass)

n_{2}=1.31 is the second medium index of refraction   (ice)

\theta_{1} is the angle of the incident ray  

\theta_{2} is the angle of the refracted ray

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

Now, the critical angle \theta_{c} is the angle from which there is no refraction and all the incident light is reflected to the same medium from which it proceeds, that is, the total internal reflection occurs. This is only possible when the index of refraction of the medium where the light strikes is higher than the index of refraction of the other medium, then the second angle (the exit angle) will reach the 90\º, for this critical incident angle \theta_{c}.

Since n_{1}>n_{1}, \theta_{1}=[tex]\theta_{c}[/tex] and \theta_{2}=90\º, hence:

n_{1}sin(\theta_{c})=n_{2}sin(90\º)

(1.90)sin(\theta_{c})=(1.31)sin(90\º)

Isolating \theta_{c}:

\theta_{c}=sin^{-1} (\frac{1.31}{1.90})

Finally:

\theta_{c}=43.5\º

The option that is close to this value is 43\º

4 0
2 years ago
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