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tatuchka [14]
3 years ago
14

if u connect 3 resistors, having values 2ohm, 3ohm, 5ohm in parallel, will the value of total resistance of will be 2ohm or grea

ter than 5ohm or lie between 2ohm and 5oh? EXPLAIN
Physics
2 answers:
Tju [1.3M]3 years ago
4 0
<span>If u want only WHether the total resistance is less than 2 or less than 5 or more than 5 ohms:  there is a Simple way.

When you connect resistances in parallel, resultant resistance is always smaller than all of them. So it is less than 2 ohms</span>.

irina1246 [14]3 years ago
4 0
\frac{1}{3}+\frac{1}{2}+\frac{1}{5}=\\&#10;\frac{10}{30}+\frac{15}{30}+\frac{6}{30}=\\&#10;\frac{31}{30}\\\\&#10;R_t=\frac{30}{31} \ \Omega\approx0.97\ \Omega
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On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question
romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(c) 45.4545 kg on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

(c) We have given that weight of the person on the earth = 100 lbs

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs \frac{100}{0.37}=270.27kg

8 0
3 years ago
If I connect an inductor (L) to a capacitor (C), I will get an LC oscillator circuit with some natural frequency omega. If I wer
mart [117]

The new natural frequency would be ω/2.

we know that,

f = \frac{1}{2 pi \sqrt{LC} } = ω.       -> equation 1

now, when capacitance is quadrupled,

f' = \frac{1}{2 pi \sqrt{L ( 4C )} }

f' = \frac{1}{2 pi (2)\sqrt{LC} }.           -> equation 2

substituting value of equation 1 in equation 2 , we get,

f' = \frac{w}{2}

Hence, the new natural frequency of the circuit is ω/2.

what do you mean by frequency ?

The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the  LC circuit's resonant frequency.

Learn more about frequency here:-

brainly.com/question/12530980

#SPJ4

8 0
2 years ago
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:
Westkost [7]

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

7 0
3 years ago
Read 2 more answers
If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W
Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

P=\frac{3600 J}{3.0 s}=1200 W

5 0
3 years ago
Read 2 more answers
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

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Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
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