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2 years ago

3

When additional resistors are added in parallel to resistors already present in a circuit the equivalent resistance of the the c

ircuit increases. (Section 19.8)
a. True.
b. False.

1 answer:

DiKsa [7]2 years ago

3 0

Answer:

This is the false statement.

Explanation:

In this statement it is said that when additional resistance added in parallel to resistors , the equivalent resistance of the circuit increases which is false statement. Because when an additional resistance is added into the system parallel to the resistors already present inside the circuit the current flow inside the circuit increases. because there are more number of branches formed inside the circuit and due to this current flow inside the circuit increases and the equivalent resistance of the circuit decreases.

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14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

Vladimir79 [104]

The relationship between resistance R and resistivity $\rho$ is
$R= \frac{\rho L}{A}$
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m$
and the cross section is
$A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2$

From the first equation, we can then find the length of the wire when $R=4.8 \Omega$ (copper resistivity: $\rho = 1.724 \cdot 10^{-8} \Omega m$ )
$L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m$

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2 years ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

7 0

2 years ago

Read 2 more answers

Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r

Helga [31]

We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

E = 1.578 x 10^ -24 J

7 0

2 years ago

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A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over

kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air $\nu\ of air =19.31*10^{-6} m2/s$

$\rho = 1.048 kg/m3$

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

$Re =\frac{VD}{\nu}$

$= \frac{ 15*0.025}{19.31*10^{-6}}$

= 1.941 *10^4

drag coffecient is given as

$C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}$

solving for f_D

$f_D = C_D A_f*\frac{\rho v^2}{2}$

$=C_D D*\frac{\rho v^2}{2}$

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

$f_D = 1.1*0.025 *\frac{1.048*15^2}{2}$

f_D = =3.24 N/m

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2 years ago

4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember

jeyben [28]

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7 m/s

Explanation:

The Doppler shift equation is given as follows;

$f' = \dfrac{v - v_o}{v + v_s} \times f$

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

$v_o$ = Observer velocity = X m/s

$v_s$ = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

$20 = \dfrac{330- v_o}{330+0} \times 21$

330 - $v_o$ = (20/21)*330

$v_o$ = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7 m/s.

8 0

2 years ago