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3 years ago

10

Our eyes perceive colors because of differences in which of the following properties of light?

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

Answer: A Wavelength

Explanation: Because it’s wavelength

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A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p

miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}$

$V_{1}=9.18\ V$

The potential on the second plate

$V_{2}=V-V_{1}$

$V_{2}=51.5 -9.18$

$V_{2}=42.32\ v$

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

$C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}$

$C=6.99\times10^{-7}\ F$

$C=0.69\ \mu F$

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0

3 years ago

What does the "c^2" mean in E=mc^2? *

Korolek [52]

Answer:

C.speed of light times 2

4 0

3 years ago

A car drives at a constant speed of 21 m/s around a circle with radius 100 m. What is the centripetal acceleration of the car?

Sauron [17]

If you look below, we are using the centripetal acceleration formula. You just need to plug your numbers in and it should work! $A_c = \frac{v^2}{r} \\A_c=\frac{21^2}{100} \\A_c=\frac{441}{100} \\A_c= 4.41m/s/s$

6 0

3 years ago

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

4 years ago

A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

<u>Now the no. of moles in this many molecules:</u>

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

7 0

3 years ago