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2 years ago

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Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

ectric field at the center of the square? (k=1/4πϵ0=8.99×109N⋅m2/C2 )
Express your answer in terms of the variables d, q, and appropriate constants.

2 answers:

Rina8888 [55]2 years ago

8 0

<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>

Vaselesa [24]2 years ago

7 0

Answer:

E = 2kq /d^2

Explanation:

Let d = side of the square

Distance between the center and corner of the square = a

a = √(d^2 + d^2) / 2

a = √2d^2 / 2

a = (√2 * d) /2

a = d /√2

Let the magnitude of an electric field = E

E = kq/a^2

Put a = d/√2

E = kq / (d/√2)^2

E = kq / d^2 / 2

E = 2kq / d^2

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An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

A)

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

2 years ago

Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po -->

GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

Atomic masses: Pb-210 = 209.98284 amu

Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

= -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

= -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

8 0

2 years ago

A submarine completed a 450 km training with an average speed of 50 km/h. For the first 180 km, it travelled at an average speed

Kryger [21]

Answer:

45km/hr

Explanation:

Total distance=450km

Total speed=50km/hr

Total time= distance/speed

=450/50

=9hrs

distance a=180km

speed a=60km/hr

Time a=180/60

=3hrs

Distance b=450-180=270km

Speed b=?

Time b=270/speed b

Total time=time a + time b

9=3+(270/speed b)

270/speed b =9-3

270/speed b =6

6*speed b =270

Speed b=270/6

Speed b=45km/hr

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2 years ago

An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin

fgiga [73]

Given that,

The electric field is given by,

$\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}$

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of $(\vec{E}\times\vec{B})$ vector is the direction of propagation of the wave .

Direction of magnetic field = $\hat{j}$

$B=B_{0}\sin(kx-\omega t)\hat{k}$

We need to calculate the poynting vector

Using formula of poynting

$\vec{S}=\dfrac{E\times B}{\mu_{0}}$

Put the value into the formula

$\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}$

$\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

Hence, The poynting vector is $\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

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2 years ago

A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters,

lina2011 [118]

496/1127 = 0.44 = 44%

<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>

<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>

<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>

<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>

3 0

2 years ago

Read 2 more answers