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Amanda [17]
3 years ago
8

Which statement correctly describes the current in a circuit that is made up of any two resistors connected in series with a bat

tery?
A. The current in the battery equals the sum of the currents in the resistors.

B. The current in the battery equals the product of the currents in the resistors.

C. The current in the battery and in each resistor is the same.

D. The current in the battery is less than the current in either resistor.
Physics
1 answer:
jeyben [28]3 years ago
6 0
The correct answer is
<span>C. The current in the battery and in each resistor is the same.

In fact, when resistors are connected in series, the current flowing through them is the same in each resistor. This is also equal to the current flowing in the circuit, so it is the same as the current flowing through the battery.</span>
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As a tornado or other storm system passes over a building, low pressure can tug a roof upward. When those forces surpass the force exerted by the weight of the roof, the structure flies up and is swept away by wind currents..

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DENIUS [597]

Answer: option d: The nucleus of Atom Q is more stable than the nucleus of Atom P.

Explanation:

Atom P is radioactive and disintegrates, it emits beta particles (high speed electrons or positrons) because it is not stable. On disintegration, it forms a stable Atom Q which is non-radioactive and thus it does not disintegrates further.

Thus, the correct option is only d. The nucleus of Atom Q is more stable than the nucleus of Atom P.

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Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha
yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}   (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m

Next, you replace the values of the parameters to calculate V:

V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV

(b) The potential electric energy is given by:

U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J

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larisa [96]

As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.

Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.

As per Planck's quantum theory -

                                                      hf =E'-E

          Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.

The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.

We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

       v=\lambda f

[here \lambda is the wavelength of the wave ]

It is given that wavelength =21 cm=0.21 m

 The frequency is calculated as -

                                                         f=\frac{v}{\lambda}

                                                                 =\frac{c}{\lambda}

                                                                 =\frac{3*10^{8} m/s }{0.21 m}

                                                                  =14.2857*10^{8} s^{-1}

 Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz

Here Hertz[ Hz] is the unit of frequency.

7 0
2 years ago
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