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Karolina [17]
3 years ago
12

How did we get 12cm ????​

Physics
2 answers:
emmainna [20.7K]3 years ago
8 0

Answer:

multiply Q d1 times 2 due to the split in the middle

Explanation:

Keith_Richards [23]3 years ago
4 0

\huge\underline\mathtt\colorbox{cyan}{Yes 12}

Explanation:

Because q is the midpoint of distance between Q and 4Q

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What does an excess of electrons produce?
MArishka [77]
Electrons are negatively charged so an excess will produce a negative charge.
4 0
3 years ago
Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.650 µC charge and flies due west at a speed of
Ganezh [65]

Answer : F=2808\times 10^{-11}\ N

Explanation :

It is given that,

Charge, q=0.650\ \mu C=0.65\times 10^{-6}\ C

Velocity of Aircraft, v=540\ m/s  (in west)

Magnetic field, B=8 \times 10^{-5}\ T ( in north )

Using the relation as :

F=q(v\times B)

Magnetic force is ,

F=0.65\times 10^{-6}\ C\times 540\ m/s\times 8 \times 10^{-5}\ T

F=2808\times 10^{-11}\ N

Using Right hand thumb rule, the direction of force is in the plane perpendicular to the velocity and the magnetic field i.e. -\hat{k}.

4 0
4 years ago
A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci
Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = \frac{C_{3}C_{4}  }{C_{3} + C_{4}  }

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0
3 years ago
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
alisha [4.7K]

Answer:

There are 4 liquids in this experiment and red is the least dense of all of them so it should float on top, which it is doing.

The red that you see at the bottom is neither liquid nor is it a part of the experiment.

It is simply the <u>color of the bottom of the container</u> that the experiment was conducted in.

4 0
3 years ago
Read 2 more answers
Which phase changes are endothermic?
konstantin123 [22]
Fusion,melting,vaporization,submlimination are examples of endothermic changes
5 0
3 years ago
Read 2 more answers
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