A series circuit contains a 20-Ω resistor, a 200-mH inductor, a 10-μF capacitor, and an ac power source. At what frequency should the power source drive the circuit in order to have maximum power transferred from the driving source?

Answers

Answer 1

Answer:

Answer:

f = 113 Hz

Explanation:

In order to have maximum power transferred from the driving source, as the RMS voltage doesn´t depends on frequency, the current I must be maximum.

This condition is met when the circuit behaves a purely resistive, as the impedance is at a minimum.

Such condition is known as resonance, and it satisfies the following equation:

XL = XC ⇒ ω₀ * L = 1 / ω₀*C, where ω₀, is the angular frequency at resonance.

Solving for ω₀, we have:

ω₀ = 1/√LC = 1/√200*10⁻3 H* 10⁻6F = 707 rad/sec

As we need to find the frequency (in cycles/sec), we need to convert from angular frequency to frequency, as follows:

ω₀ = 2*π*f₀ ⇒ f₀ = ω₀ / 2*π = 707 rad/sec / 2*π rad = 113 Hz

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Answers

Answer:

Explanation:

A rectifier is an electrical device that converts alternating current (AC) to direct current (DC), a process known as rectification. Rectifiers have many uses including as components of power supplies and as amplitude modulation detectors (envelope detectors) of radio signals. Rectifiers are most commonly made using solid state diodes but other type of components can be used when very high voltages or currents are involved. When only a single diode is used to rectify AC (by blocking the negative or positive portion of the waveform), the difference between the term diode and the term rectifier is simply one of usage. The term rectifier describes a diode that is being used to convert AC to DC. Most rectifier circuits contain a number of diodes in a specific arrangement to more efficiently convert AC power to DC power than is possible with only a single diode.

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact

Answers

Answer:

$y_(max) = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,(m)/(s))^(2) + 2\cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m)}$

$v = 2.913\,(m)/(s)$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m) + (1)/(2)\cdot (3\,kg)\cdot (2.913\,(m)/(s) )^(2) = (3\,kg) \cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s) ^(2)$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = (1)/(2)\cdot (2000\,(N)/(m) )\cdot (\Delta s)^(2) - (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot \Delta s$

$1000\cdot (\Delta s)^(2)-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_(1) \approx 0.128\,m$

$\Delta s_(2) \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s)^(2) = (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot y_(max)$

$y_(max) = 0.829\,m$

Answer:

0.81 m

Explanation:

In all moment, the total energy is constant:

Energy of sistem = kinetics energy + potencial energy = CONSTANT

So, it doesn't matter what happens when the block hit the spring, what matters are the (1) and (2) states:

(1): metal block to 0.8 m above the floor

(2): metal block above the floor, with zero velocity ( how high, is the X)

Then:

$E_(kb1) + E_(gb1) = E_(kb2) + E_(gs2)$

$E_(kb1) + E_(gb1) = 0 + E_(gs2)$

$(1)/(2)*m*V_(b1) ^(2) + m*g*H_(b1) = m*g*H_(b2)$

$H_(b2) = (V_(b1) ^(2) )/(2g) + H_(b1)$

Replacing data:

$H_(b2) = (0.44^(2) )/(2*9.81) + 0.8$

$H_(b2) = (0.44^(2) )/(2*9.81) + 0.4$

HB2 ≈ 0.81 m

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Answers

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire

Answers

Answer:

118 m/s

Explanation:

Given :

We know that

$f\ =\ (1)/(2l) \sqrt{(T)/(U) }$ ......Eq(1)

Where $\sqrt{(T)/(u) }$ =v

l=length

f=frequency

l= 98.0 cm= 0.98 m

f=60.0 Hz

Now from the Eq(1)

$f\ =\ (v)/(2l)$

This equation can be written as

v=2fl.............Eq(2)

Putting the value f and l in Eq(2)

v=2*60*0.98

v=117.6 m/s ~ 118 m/s

WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”picture attached

Answers

Answer:

Explanation:

B is the answer sorry for the late response

Look at the figure below and calculate the length of side y.A. 8.5
B. 17
12
y
C. 6
O D. 12
45
Х

Answers

Answer:

I want to say a because you want to subtract and simplify

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