A cup has a mass of 0.0650 kg and a

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

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

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

5 0

3 years ago

What is projectile motion caused by?

Lemur [1.5K]

Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

6 0

3 years ago

Read 2 more answers

Models are frequently used in science to assist in the understanding of complex information. Models can include items like a wor

nordsb [41]

B. Three dimensional I’m almost positive

3 0

3 years ago

A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar ec

Sonja [21]

1) A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.

2) anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.

3) during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA.

4) a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.

5) a partial lunar eclipse begins when the moon first touches earth's -->
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.

6) a point at which the moon crosses earth's orbital plane is called a(n) -->
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".

5 0

3 years ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

8 0

3 years ago