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AlladinOne [14]
3 years ago
7

A coil with internal resistance can be modeled as a resistor and an ideal inductor in series. Assume that the coil has an intern

al resistance of 2.00 omega and an inductance of 480 mH. A 1.60 1.60 MuF capacitor is charged to 24.0 V and is then connected across the coil.
(a) What is the initial voltage across the coil? V
(b) How much energy is dissipated n the circuit before the oscillations die out? mJ
(c) What is the frequency of oscillation of the circuit? (Assume the internal resistance is sufficiently small that it has no impact on the frequency of the circuit.) Hz
(d) What is the quality factor of the circuit?
Physics
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:The impedance per phase is 1.25H. This was calculated using the ... the complex power generated is 3.4kw. The full solution can be found below.

Explanation:

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Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s
Ne4ueva [31]
The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
F=qvB \sin \theta
where \theta is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is q=2.7 \mu C=2.7 \cdot 10^{-6}C. In this case, v and B are perpendicular, so \theta=90^{\circ}, therefore we have:
B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T

2) In this second case, the angle between v and B is \theta=55^{\circ}. The charge is now q=42.0 \mu C=42.0 \cdot 10^{-6}C, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N
5 0
3 years ago
Read 2 more answers
The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
3 years ago
The ancient Egyptians used the stars to _____.
OleMash [197]

Answer:

predict annual flooding

4 0
2 years ago
Read 2 more answers
Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

7 0
3 years ago
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0
2 years ago
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