Which elements are metalloids

12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L

Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0

3 years ago

Define friction State the law of frictionGive two factors that may affect friction

umka21 [38]

Friction:

Friction is the resistive force that acts on the body in a direction opposite to the direction of motion of the same object.

Laws of friction:

1. The frictional force acting on the object is proportional to the normal force acting on the object.

For example, if the object of mass m is moving along the horizontal surface, the normal force acting on the object is mg.

Thus, the frictional force acting on the object is,

$\begin{gathered} F_r\propto F_N \\ F_r=\mu F_N \end{gathered}$

where F_N is the normal force acting on the object,

$\mu\text{ is the coefficient of friction}$

The diagrammatic representation of the frictional force acting on the object is,

2. The frictional force acting on the object depends upon the nature of the surface in contact with the object.

3. The frictional force does not depend upon the area of contact of an object with the surface.

4. The kinetic friction acting on the object is independent of the velocity of the object.

5. The coefficient of static friction is more than the coefficient of kinetic friction.

Two factors on which the frictional force depends are:

1. The frictional force depends on the adhesion between the two bodies in contact.

2. The frictional force acting on the depends upon the roughness of the surface on which the object is moving.

3 0

1 year ago

Think about the various energy sources such as the sun, fossil fuels, water, wind, and nuclear power. What is the energy source

lakkis [162]

Answer:

The energy source that powers my home is gotten from burning of fossil fuels.

No! I do not like this energy source.

I personally would prefer solar source of electricity

Explanation:

Fossil fuels are fuels gotten from the decomposition of dead organisms over time, under intense heat and pressure. They are usually found buried beneath the earth's crust where they have been formed and trapped.

Most electricity generating stations generate electricity by burning fossil fuels like natural gas and gasoline to generate electricity. The problem with fossil fuels are the various part that they play in increasing carbon footprints in the atmosphere. The excess carbon in the atmosphere has been a major contributor to the global warming of planet Earth.

My preference for solar energy source is first due to its abundance unlike the fossil fuels that are already diminishing in storage beneath the earth's crust. Also, solar energy is a clean source of energy that does not leave any damage on earth from its use. It also promises to be a cheap source of power in the future with advances in solar technologies.

8 0

3 years ago

"5 N, up" is an example of a ___. OA) force OB) mass OC) weight OD) magnitude

Sunny_sXe [5.5K]

Answer:

A) Force

Explanation:

It is an example of force since force is a vector quantity so it has magnitude and direction. In this case the magnitude is equal to 5 [N] and the direction is upward.

The weight can not be, as it always acts downward.

Mass is not a force, its unit is given usually in kilogram [kg]

5 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago