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MAVERICK [17]
3 years ago
15

Which of the following codes is used to report the removal of 37 skin tags by electrosurgical destruction?

Physics
1 answer:
Tema [17]3 years ago
7 0
Are there any answer choices??
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Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
3 years ago
The mass of an electron is ______
Over [174]
I believe that the answer would be atomic mass.<span />
5 0
3 years ago
Read 2 more answers
A railroad car having a mass of 17.5 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass
Leokris [45]

Answer:

Then final velocity of the coupled system of cars will be 0.58 m/s

Explanation: Let us consider the car-1 is moving towards right (say + ve direction) and car-2 is moving towards left (say in - ve direction), accordingly velocity are considered +ve and -ve.

m_{1} = 17.5 \ Mg = 17.5 \times 10^{6} \ g = 17.5 \times 10^{3} \ Kg = 17500 \ Kg

v_{1} = + \ 1.5 \ m/s

m_{2} = 12 \ Mg = 12 \times 10^{6} \ g = 12 \times 10^{3} \ Kg = 12000 \ Kg

v_{2} = - \ 0.75 \ m/s

Applying the conservation of momentum, and let the final velocity of combined system is V m/s

m_{1} \times v_{1} + m_{2} \times v_{2} = (m_{1} + m_{2}) \times V

17500 \times 1.5 \  + 12000 \times (-0.75) = 29500 \times V

26250 - 9000 = 29500 \times V

V = \frac{17250}{29500} = 0.58 m/s

Then final velocity of the coupled system of cars will be 0.58 m/s

8 0
3 years ago
Read 2 more answers
Sorry I keep asking questions, but if two forces on an object are balanced, will it move? (Im also trying to get rid of my point
Dmitrij [34]
If they are equal amounts in force then No it won't.
6 0
3 years ago
Read 2 more answers
A coil is connected in series with a 19.0 kΩ resistor. An ideal 50.0 V battery is applied across the two devices, and the curren
Gre4nikov [31]

Answer:

inductance of the coil 29.3767  H

Explanation:

given data

resistor R =  19.0 kΩ = 19 × 10³ Ω

potential applied V  = 50.0 V

current I = 2.20 mA  =  2.20 × 10^{-3} A

time t = 2.80 ms = 2.80 × 10^{-3} s

solution

we know for maximum current in circuit that is

current = V ÷ R   .........1

current =  \frac{50}{19\times 10^3}

current = 2.63 × 10^{-3} A

so at time t = 0

t = -\frac{L}{R} ln(1-\frac{I_f}{I_{max}})  

2.80 \times 10^{-3} = -\frac{L}{19\times 10^3} ln(1-\frac{2.20\times 10^{-3}}{2.63\times 10^{-3}}})  

L = 29.3767

3 0
3 years ago
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