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sertanlavr [38]

2 years ago

6

Two resistors of 12Ω and 2Ω are connected in parallel to a battery of 4 V. Current through 12Ω resistor will be _____________ an

d power dissipated at 2Ω resistor will be ??

1 answer:

Alex2 years ago

6 0

Answer:

0.33 Amp

8 Watts

Explanation:

The current through the 12 Ω resistor will be given by Ohm's law:

4 V / 12 Ω = 1/3 Amp ≈ 0.33 Amp

The power dissipated on the 2 Ω resistor can be calculated via the formula: $\frac{V^2}{R}$ which gives:

4^2 / 2 = 8 Watts

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lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

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Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

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If you wanted to discover the youngest stars you could find in some grouping of stars in the Galaxy, which type of star group wo

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Answer:

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$v_{f}$ = Final speed of rock = 392 ms⁻¹

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Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

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