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19 days ago

15

A uniformly charged rod (length = 2.0 m, charge per unit length = 5.0 nc/m) is bent to form one quadrant of a circle. what is th

e magnitude of the electric field at the center of the circle?

1 answer:

inna [2.2K]19 days ago

4 0

The electric field at the center of a circular arc can be determined using the formula

$E = \frac{2k\lambda sin\frac{\theta}{2}}{R}$

and we know that

$\lambda = 5nC/m$

$k = 9 \times 10^9 Nm^2/C^2$

$\frac{\pi}{2}R = L = 2m$

$R = \frac{4}{\pi}$

furthermore, we are aware that

$\theta = \frac{\pi}{2}$

now applying the formula above

$E = \frac{2(9\times 10^{-9})(5\times 10^{-9})sin45}{4/\pi}$

$E = 50 N/C$

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The magnetic field is calculated to be -6.137 × T. Explanation: Given the radio wave wavelength of λ = 0.3 m and an intensity of I = 45 W/m² at times t = 0 and t = 1.5 ns, we determine Bz at the origin. We use the intensity formula relating to the electric field, which incorporates the known intensity of 45, the speed of light c = 3 × m/s², and ∈o as 8.85 × C²/N.m², leading us to E = 184.15. Consequently, applying the equations, we find B = -6.137 × T at the z-axis.

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4 days ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

Softa [2029]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Charge (Q₁) = 572.8 $* 10^{-9}$ C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

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25 days ago

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

serg [2593]

Answer:

$v = 3369.2 m/s$

Explanation:

The beacon is rotating at an angular speed of

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

We know that

$v = r \omega$

At this point we have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

So we can conclude with

$v = 3218(1.047)$

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19 days ago

A 6V radio with a current of 2A is turned on for 5 minutes. Calculate the energy transferred in joules

ValentinkaMS [2425]

Answer:

R=V/I=6/2=3 ohm

time = 5 minutes = 5*60=300 seconds

I=2 A

Energy = I²Rt=(2)²*3*300=4*900=3600 J

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1 month ago

The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.

Maru [2355]

Answer:

11109.825 N

Explanation:

Provided Information:

mass = m = 1510 kg

initial acceleration (a) = 0.75g (where g = 9.81 m/s²)

Using the formula F=ma

= (1510)*(0.75*9.81)

= 11109.825 N

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19 days ago