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3 years ago

Is kcl and kvl applicable to ac circuit?

Physics

1 answer:

Sedaia [141]3 years ago

8 0

Yes, they both are, and you'd just be working with complex numbers instead of real number.

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Which of the following places would u expect to have the warmest climate ?

adelina 88 [10]

B. Location#2 with an altitude of 200 feet

7 0

3 years ago

a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this

Yuri [45]

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

4 0

3 years ago

Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measure

chubhunter [2.5K]

Answer:

the smallest angle from the antennas is 47.3°

Explanation:

We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.

Therefore, the relation is:

λ = c /f

where

c is the speed of light constant
λ is the wavelength
f is the frequency

Thus,

λ = (3 × 10⁸ m/s) / (3.4 MHz)

= (3 × 10⁸ m/s) / (3.4 MHz)(10⁶ Hz/1 MHz)

= 88.235 m

Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as

θ = sin⁻¹(λ / d)

where

d is the distance between the two radio antennas

Thus,

θ = sin⁻¹(88.235 / 120)

θ = 47.3 °

Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is 47.3 °.

5 0

3 years ago

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

Minchanka [31]

The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression.
In general, the potential energy of gravitational field is defined as:
$U=-G \frac{mM}{r}$
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
$F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}$
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
$g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg$
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
$U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}$
When we plug in the numbers we get:
$U=-4.99\cdot 10^{10} J$
The potential energy has to be equal to the kinetic energy.
$E_k=4.99\cdot 10^{10} J$

3 0

3 years ago

How long will it take a car to go from a complete stop to 44km/h if they are accelerating at 5km/h

Ber [7]

Explanation:As always, if the relation does not come to mind immediately, take a glance at the basic kinematic equations of constant acceleration

what do we know? Vi start speed; Vf terminal speed; acceleration a; and we want to know time t. Three of the relations have a term in d distance - which we don’’t know. That leaves this one: Vt = Vi + a*t. Here we go then:

44 = 0 + 5 * t so t = 44 / 5 = 8.8 hours

4 0

3 years ago