How many hours are in a month of 30 days?

For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr

Marina CMI [18]

Answer:

L= 2 mH

$C=1.26\times 10^{-7}\ F$

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

$E=\dfrac{1}{2}LI^2$

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

$10^{-5}=\dfrac{1}{2}\times L\times 0.1^2$

$L=\dfrac{2\times 10^{-5}}{0.1^2}\ H$

L=0.002 H

L= 2 mH

We know that frequency f is given as

$2\pi f=\dfrac{1}{\sqrt{LC}}$

C=Capacitance , f=frequency ,L=Inductance

Now by putting the values

$2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }$

$62831.85=\dfrac{1}{\sqrt{0.002\times C}}$

$\sqrt{0.002\times C=\dfrac{1}{62831.85}$

$0.002\times C=0.0000159^2$

$C=\dfrac{0.0000159^2}{0.002}\ F$

$C=1.26\times 10^{-7}\ F$

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0

3 years ago

Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet

Valentin [98]

Answer:

v = 8.09 m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

starting point. Higher

Em₀ = U = m gh

final point. To go down the slope

Em_f = K = ½ m v²

The work of the friction force is

W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

N - W_y = 0

N = W_y

X axis

Wₓ - fr = ma

let's use trigonometry

sin θ = y / L

sin θ = 11/110 = 0.1

θ = sin⁻¹ 0.1

θ = 5.74º

sin 5.74 = Wₓ / W

cos 5.74 = W_y / W

Wₓ = W sin 5.74

W_y = W cos 5.74

the formula for the friction force is

fr = μ N

fr = μ W cos θ

Work is friction force is

W_fr = - μ W L cos θ

Let's use the relationship of work with energy

W + ΔU = ΔK

-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²

v² = - 2 μ g L cos 5.74 +2 (gh)

v² = 2gh - 2 μ gL cos 5.74

let's calculate

v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

v² = 215.6 -150.16

v = √65.44

v = 8.09 m/s

6 0

3 years ago

If these gas molecules were compressed further, how would the average kinetic energy of the molecules be affected? A. It would i

Advocard [28]

I believe the correct response would be B. It would decrease.

8 0

3 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from

Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0

3 years ago

Read 2 more answers

If an object has a mass of 38 kg, what is its approximate weight on earth?

klasskru [66]

38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.

3 0

3 years ago