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larisa86 [58]
3 years ago
12

How many electrons must be removed from an electrically neutral silver dollar to give it a charge of +2.4 μC?

Physics
1 answer:
sasho [114]3 years ago
3 0
The total charge to be removed is -<span>2.4 μC</span>. The number of
electrons corresponding to this charge is
                      N = (-2.4 x 10^-6 C)/(-1.60 x 10^-19 C)
                          = 1.5 x 10^13 electrons
Therefore 1.5 x 10^13 electrons need to be removed from the neutral silver dollar.
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Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,
Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

 

6 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

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A, electromagnetic radiation
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What is described by the terms body-centered cubic and face-centered cubic?
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In physical chemistry, the terms body-centered cubic  (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.

It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.

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The image shows a pendulum that is released from rest at point A. Shari tells her friend that no energy transformation occurs as
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