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3 years ago

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Physics

1 answer:

katrin [286]3 years ago

4 0

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

$E = \dfrac{kq}{r^2}$

k is the coulomb constant

$q= \dfrac{Er^2}{k}$

$q= \dfrac{1.18\times 0.822^2}{9\times 10^9}$

q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

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What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time?

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Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant

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3 years ago

Trên bóng đèn có ghi là 12v -6w. Khi đèn sáng bình thường thì dòng điện chạy qua đèn có cường độ là

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Answer:

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3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

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3 years ago

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3 years ago

A solenoid of length 5.0 cm has a cross sectional area of 1.0 cm2 and has 300 uniformly spaced turns. The solenoid is wrapped in

OleMash [197]

The calculated mutual inductance is 8.544 x 10⁻⁵ H.

Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.

Length of the solenoid= 5.0 cm

Area of cross-section=1.0 cm²

no of spaced turns=300 turns

turns of insulated wire=180 turns

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=79.12 x 10⁻¹⁰ x 54000 / 5

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hence, the mutual inductance is 8.544 x 10⁻⁵ H.

Learn more about Mutual inductance here-

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1 year ago