CAN SOMEONE PLEASE HELP ME!!!!! IM BEGGING YOU!!

HURRY I NEED THIS ASAP

gizmo_the_mogwai [7]

Answer:

D

data as proportions of a whole

8 0

3 years ago

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An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00

Basile [38]

Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

$V=initial\ rate\times inductance$

$initial\ rate=\dfrac{V}{L}$

Put the value into the formula

$initial \rate=\dfrac{6.00}{2.30}$

$initial\ rate=2.60\ A/s$

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

$rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}$ ....(I)

Where, $T=\dfrac{L}{R}$

$T=\dfrac{2.30}{8.00}$

$T=0.2875$

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

$e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})$

$e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})$

$e^{\dfrac{-t}{T}}=0.333$

The rate of current increases is

Put the value in the equation (I)

$rate\ of\ current\ increase=2.60\times0.333$

$rate\ of\ current\ increase=0.8658\ A/s$

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

Put the value into the formula

$i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})$

$i=0.435\ A$

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

$i=\dfrac{V}{R}$

$i=\dfrac{6.00}{8.00}$

$i=0.75\ A$

The final steady-state current is 0.75 A.

Hence, This is the required solution.

3 0

3 years ago

How strong is the attractive force between a glass rod with a 0.700μC0.700μC charge and a silk cloth with a –0.600μC–0.600μC cha

lisov135 [29]

Answer:

0.2625 N/C

Explanation:

Force of attraction = $\frac{K\timesQ_1\times Q_2}{R^2}$

where Q₁ and Q₂ are charges and R is distance between charges.K is a constant equal to 9 x 10⁹.

= $\frac{9\times10^9\times.7\times10^{-6}\times.6\times10^{-6}}{(12\times10^{-2})^2}$

= 0.2625 N/C

4 0

3 years ago

The graph below shows how the speed of a bus change during part of a journey. Use this graph to answer questions 3-5

Minchanka [31]

Answer:

O to A and D to E are accelorating.

Explanation:

You can tell because the graphed points have positive slopes in these locations.

6 0

3 years ago

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Four charges are located at the corners of a square. Each side of the square is 37 m. The left two charges have a positive charg

nadezda [96]

Answer:

E = 781.12 N/C

Explanation:

Look at the attached graphic:

The 20µC charges are positive , then, the electric fields leave the charge.

The 22µC charges are negative, then, the electric fields enter the charge.

The electric field due to each of the charges is calculated by Coulomb's law:

E= k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Problem development

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

E₁, E₂, E₃, E₄: Electric field at point P due to charge q₁, q₂, q₃, and q₄ respectively

The electric field in the direction of the y axis at the point P in the center of the square is equal to zero because the sum of the vertical components upwards is equal to the sum of the vertical components downwards.

Calculation of the electric field in the direction of the x-axis in the center of the square (point P)

Eₓ = E₁ₓ + E₂ₓ + E₃ₓ + E₄ₓ

$d = \sqrt{18.5^2+ 18.5^2} =26.16m$

E₁ₓ = E₂ₓ = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(26.16²) = 185.98 N/C

E₃ₓ = E₄ₓ = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(26.16²) = 204.58 N/C

Eₓ = E = 2*185.98 +2*204.58 = 781.12 N/C

5 0

3 years ago