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3 years ago

What is the equivalent resistance if you connect three 10.0 Ω resistors in series?

Physics

2 answers:

Sonja [21]3 years ago

4 0

Req = 30.0Ω.

When two or more resistors are in series, the intensity of current that passes through each of them is the same. Therefore, if you notice, you can observe that the three previous series resistors are equivalent to a single resistance whose value is the sum of each one.

Req = R1 + R2 + R3 = 10.0Ω + 10.0Ω + 10.0Ω = 30.0Ω

denpristay [2]3 years ago

4 0

Answer:

Req = 30.0Ω.

Explanation:

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The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that

OLEGan [10]

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

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2 years ago

Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p

irina1246 [14]

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = $\frac{4}{2}$ =2ohm [ proven as in series $R_{e} = 2 + 2 = 4ohm$ ]

to calculate the equivalent resistance when in parallel:

$\frac{1}{R_{e} } = \frac{1}{R_{1} } + \frac{1}{R_{2}}$

so,

$\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}$

$R_{e} = 1ohm$

4 0

2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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Answer:

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