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Ad libitum [116K]
3 years ago
6

Three resistors (16 ohm, 16 ohm and 8 ohm) are connected in parallel. The equivalent

Physics
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

4 ohm

Explanation:

The equivalent resistance (Re) of three resistors in parallel is given by;

1/Re = 1/R1 + 1/R2 + 1/R3

Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm

1/Re= 1/16 + 1/16 + 1/8

1/Re= (0.0625) + (0.0625) + (0.125)

Re= 4 ohm

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In an experiment you measure a first-order red line for Hydrogen at an angle difference of ΔΘ = 22.78o. The diffraction grating
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Answer:

a) wavelength = 656.3 nm

b)  the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

Explanation:

Given that;

angle of diffraction Θₓ = 22.78°

incident angle Θ₁ = 0

slit separation d  = 5900 lines per cm = 1/5900 cm = 10⁻²/5900 m = 0.01/5900 m

order of diffraction n = 1

wavelength λ = ?

to find the wavelength, we use the expression

λ = d (sinΘ₁ + sinΘₓ) / n

To find the wavelength λ;

λ = 0.01/5900 × (sin0 + sin22.78° )

λ = 6.5626 × 10⁻⁷ m

λ = 656.3 x 10⁻⁹ m

∴ λ = 656.3 nm

b)

According Balnur's  series spectral lines; n₁ = 3, n₂ = 2 and

λ = R [ 1/n₂² - 1/n₁²]

where  R is Rydberg's constant

from λ = R [ 1/n₂² - 1/n₁²]

R = 1/λ [n₂²n₁² / n₁² - n₂²]

R = 10⁹/ 656.3 [ 9 × 4 / 9 - 4 ]

R = 1.097 × 10⁷ m⁻¹

Therefore the value of Rydberg's constant for this measurement is 1.097 × 10⁷ m⁻¹

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Science and technology are closely related, but not the same. Which of the following is most likely a result of technology, not
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Explanation:

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Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge
MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

V_1 = V_2 (1)

Keeping in mind the relationship between charge, voltage and capacitance:

C=\frac{Q}{V}

we can re-write (1) as

\frac{Q_1}{C_1}=\frac{Q_2}{C_2} (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

C=4 \pi \epsilon_0 R

where R is the radius of the sphere. Substituting this into (2), we find

\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2} (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

R_2 = 2R_1

So the eq.(3) becomes

\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}

And re-arranging it we find:

Q_2 = 2Q_1

And since we know that the total charge is

Q_1 + Q_2 = 7.00 nC

we find

Q_1 = 2.33 nC\\Q_2 = 4.67 nC

3 0
4 years ago
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