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2 years ago

11

An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6

kg⋅m2.What is the magnitude of the angular acceleration of the object?
[Please answer if you know]

1 answer:

Korolek [52]2 years ago

3 0

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law. Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

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3 years ago

For the circuit shown, R = 75.0 ohms, L= 55.0 mg, and C = 25.0 μC. The power source has 12.0 V arms and a frequency of 60.0 Hz.

Natasha_Volkova [10]

Explanation:

Given that,

Resistance R = 75.0 ohms

Inductance L = 55.0 mH

Capacitance $C = 25.0\ \mu C$

Voltage V = 12.0 V

Frequency f = 60.0 Hz

We need to calculate the angular frequency

Using formula of angular frequency

$\omega = 2\pi f$

Put the value into the formula

$\omega =2\times3.14\times60.0$

$\omega=376.8\ rad/s$

(a). We need to calculate the value of $X_{L}$

Using formula of $X_{L}$

$X_{L}=\omega\times L$

Put the value into the formula

$X_{L}=376.8\times55.0\times10^{-3}$

$X_{L}=20.724\ \Omega$

(b). We need to calculate the value of $X_{L}$

Using formula of $X_{C}$

$X_{C}=\dfrac{1}{\omega C}$

$X_{C}=\dfrac{1}{376.8\times25.0\times10^{-6}}$

$X_{C}=106.16\ \Omega$

(c). We need to calculate the value of Z

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

Put the value into the formula

$Z=\sqrt{75.0^2+(20.724-106.16)^2}$

$Z=113.68\ \Omega$

(d). We need to calculate the rms current

Firstly we need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{12.0}{75.0}$

$I=0.16\ A$

Using formula of rms current

$I_{rms}=\dfrac{I_{0}}{\sqrt{2}}$

$I_{rms}=\dfrac{0.16}{\sqrt{2}}$

$I_{rms}=0.113\ A$

(e). We need to calculate the rms voltage across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

$V_{rms}=0.113\times75.0$

$V_{rms}=8.475\ V$

(f). We need to calculate the rms voltage across the inductor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{L}$

$V_{rms}=0.113\times20.724$

$V_{rms}=2.342\ V$

(g). We need to calculate the rms voltage across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.113\times106.16$

$V_{rms}=11.99\ V$

(h). We need to calculate the dissipated power by the circuit

Using formula of dissipated power

$P=RI^2$

Put the value into the formula

$P=75.0\times0.113^2$

$P=0.958\ W$

Hence, This is the required solution.

3 0

3 years ago