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3 years ago

All electric circuits have devices that are run by _____ energy.

Physics

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer: electrical energy

Explanation:

Electrical energy

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Light of frequency 5*10^14hz liberates electron with energy 2.31*10^-19 joule from a certain surface .what is wave length of ult

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Answer: Light of frequency 5 x 1014 HZ liberates electrons with energy 2.3 x 10-19from a certain metallic surface.

Explanation:

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1 year ago

Momentum is a vector quantity that depends

MissTica

On the change in potential energy

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3 years ago

In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p

UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

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2 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2

butalik [34]

Answer:

Number of turns in secondary will be 7

Explanation:

We have given primary voltage $V_p=120volt$

Number of turns in the primary is $N_p=3575$

Secondary voltage is given $V_s=235mV=0.235volt$

We have to find the number of turns in secondary

We know that $\frac{N_p}{N_s}=\frac{V_p}{V_s}$

So $\frac{3575}{N_s}=\frac{120}{0.235}$

$N_s=6.60$

As the number of turns can not be in decimal so number of turns will be 7

6 0

3 years ago