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3 years ago

10

Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

in series with a 20e-2 F capacitor, a 20-mH inductor, and a 50- resistor. -1.37 radians 0.134 radians 0.0180 radians 0.0300 radians

1 answer:

Vinvika [58]3 years ago

3 0

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

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2 years ago

allochka39001 [22]

Answer:

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2 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

approximately <u>9.18°</u>.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

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6 0

2 years ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

3 years ago

Whenever an object exerts a force on another object, the second object exerts a force o the same amount, but in the ______ direc

Brut [27]

Answer:

Opposite

Explanation:

Newton's third law of motion states that for every action there is an equal but opposite reaction.

Action-reaction force pairs make it possible for fishes to swim, birds to fly, cars to move etc,

For example, while driving down the road, a firefly strikes the windshield of a car (Action) and makes a quite obvious mess in front of the face of the driver (Reaction) i.e the firefly hit the car and the car hits the firefly.

The ultimately implies that, in every interaction, there is a pair of equal but opposite forces acting on the two interacting physical objects.

Hence, whenever any physical object exerts a force (action) on another physical object, the second physical object exerts a force (reaction) of the same amount, but acting in opposite direction to that of the first physical object.

4 0

3 years ago