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3 years ago

A capacitance C and an inductance L are operated at the same angular frequency.

Physics

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer:

(B) 7279.70 rad/sec (C) $X_L=37.1265\ ohm\ and \ X_C=37.1265\ ohm$

Explanation:

We have given $L=5.10mH$ and $C=3.70\mu F$

(B) The angular frequency is denoted as $\omega$ whose value is given by $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.10\times 10^{-3}\times 3.70\times 10^{-6}}}=7279.70\ rad/sec$

Capacitive reactance $X_C=\frac{1}{\omega C}=\frac{1}{7279.70\times 3.70\times 10^{-6}}=37.1265\ ohm$

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7Which of the following terms describes how glaciers move?

tatuchka [14]

Answer:

D is the answer I think (0 w 0 )

Explanation:

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3 years ago

Read 2 more answers

The number of confirmed exoplanets is (a) less than 10; (b) roughly 50; (c) more than 500; (d) more than 5000.

Bad White [126]

Answer:

Explanation:

Until 2019, more than 3000 planetary systems have been discovered that contain more than 4000 exoplanets, since some of these systems contain multiple planets. Most known extrasolar planets are gas giants equal to or more massive than the planet Jupiter, with orbits very close to its star.

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3 years ago

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w

lilavasa [31]

Answer:

Mutual inductance, $M=2.28\times 10^{-5}\ H$

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

$M=\dfrac{\mu_oN_1N_2A}{2\pi r}$

(b) It is given that,

$N_1=550$

$N_2=290$

Radius, r = 10.6 cm = 0.106 m

Area of toroid, $A=0.76\ cm^2=7.6\times 10^{-5}\ m^2$

Mutual inductance, $M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}$

$M=0.0000228\ H$

$M=2.28\times 10^{-5}\ H$

So, the value of mutual inductance of the toroidal solenoid is $2.28\times 10^{-5}\ H$ . Hence, this is the required solution.

8 0

3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago

A man paddles a canoe at 6 km per hour. If he paddles on a river with a current of 6 km per hour, what is the speed of the canoe

Romashka-Z-Leto [24]

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

$Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]$

5 0

3 years ago