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3 years ago

Which resistors in the circuit must always have the same current?

Physics

2 answers:

yuradex [85]3 years ago

7 0

Answer:

it's a and d

Explanation:

because it all goes around if that makes sense.. and if u guys are here from apex

kakasveta [241]3 years ago

4 0

Answer:

Resistors in series in the circuit must always have the same current

Explanation:

Resistors are said to be connected in series if they are connected one after another.

The total resistance in the circuit with resistors connected in series is equal to the sum of individual resistances.

Individual resistors in series do not get the total source voltage. Total source voltage divide among them.

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In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

3 years ago

Any collection of things that have some influence on one another can be thought of as a system. When we talk about a system, we

Katen [24]

Answer:

Explanation:

Its B

6 0

3 years ago

Helen [10]

Answer:

wavelength decreases and frequency increase

Explanation:

the higher the wavelength the smaller the frequency , the smaller the wavelength the higher the frequency

5 0

3 years ago

Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun

Nookie1986 [14]

Answer:

The answer is " $3.83 \times 10^9 \ m$ "

Explanation:

Z=2, so the equation is $E= \frac{-4B}{n^2}$

Calculate the value for E when:

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies

In this transition, the wavelength of the photon emitted is:

$\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})$

$\lambda = \frac{h c}{\Delta E}$

$h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\$

6 0

3 years ago