I need help!!! I need to have pictures for Absolute and Apparent Magnitude! Someone please help?!!!

How to find i1, i2,i3

MrRissso [65]

to find i1, i2, and i3 we need to find the total current.

to find the total current, you need to find the total resistance

you're already given the total voltage, Vs

to find Rtotal, start from the resistors furthest from the voltage source.

R3 and R4 are in series so

Rtotal= R3+R4 = 6+3 = 9 ohms

9 ohms is now in parallel with R2 so,

Rtotal= $(\frac{1}{R3+R4}) ^{-1}\\$ + (\frac{1}{R2}) ^{-1})^-1= (1/18)^-1 +( 1/9)^-1 = 6 ohms

6 ohms is in series with R1 so

Rtotal= 4+6=10 ohms

itotal= (\frac{Vtotal}{Rtotal})

= 120 v/10 ohms = 12 A

i total = i1 because all the current flows through it

i1= 12A

so the current splits into i2 and i3 and the amount of current that flows through a branch depends on the total resistance in each branch.

we already calculated the resistance in the R3+R4 & R2 branch as 6 ohms

since r3 and r4 are in series, the same current will flow through them

r3+r4 = 9 ohms

r2= 18 ohms

so the current in r2 will be half that of R3 & R4 (V=IR)

using the current divider rule

$Ix = Itotal * \frac{Rtotal}{Rx}$

i2= 12A x (6 ohms/18 ohms)= 4 A

i3= 12A x (6 ohms/9 ohms) = 8 ohms

6 0

3 years ago

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 equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

3 years ago

Need help ASAP please and thank you

SashulF [63]

Answer:

C

Explanation:

Im not sure but I did somthing simalier

8 0

3 years ago

Read 2 more answers

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$