Physics Middle School

In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?

In The Diagram, Q1 = +6.60*10^-9 C And Q2 = +3.10*10^-9 C. Find The Magnitude Of The Total Electric Field

Answers

Answer 1

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

Answer 2

Answer:

1258.46 N/C

Explanation:

acellus

Related Questions

What did Thomson’s and Rutherford’s experiments have in common? They both used charged particles in their experiments. They both used emission spectra to discover electron clouds. They both compared the atom to an indivisible sphere. They both discovered electron clouds.

Answers

Answer:

Both Thomson and Rutherford used charged particles in their experiments.

Explanation:

Theo both used charged particles

What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)
at 308 K? Use the equation: -mv2
2
For m, use 0.02000 kg. Remember that R = 8.31 J/(mol-K).
= ER
3
2
nRT
A. 1540 m/s
B. 876 m/s
C. 87.6 m/s
O D. 15,400 m/s

Answers

Answer:

v = 876 m/s

Explanation:

It is given that,

Number of mol of Neon is 2 mol

Temperature, T = 308 K

Mass, m = 0.02 kg

Value of R - 8.31 J/mol-K

We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,

[tex]\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s[/tex]

So, the correct option is (B).

I think it is 887m/s I hope this helps if not I’m really sry

Find the x-component of this vector: 47.3 m 39.4 ° Remember, angles are measured from the + x axis. x-component (m)

Answers

The x-component of the vector is 36.55 m.

Components of vectors

The components of a vector include both x and y components. The x-component is determined along x-axis, while y-component is determined along y-axis.

Vx = Vcosθ

Vy = Vsinθ

where;

Vx is the x-component of the vectorVy is the y-component of the vector

X-components of the vector

The x-component of the vector is calculated as follows;

x = 47.3 x cos(39.4)

x = 36.55 m

Thus, the x-component of the vector is 36.55 m.

Learn more about x-component of vector here: https://brainly.com/question/13416288

fill in the term to complete each sentence

a wave carries___ from one place to another.
mechanical waves carry energy through___.

Answers

Energy

A material medium

A wave carries energy from one place to another. Mechanical waves carry energy through a material medium.

What is a wave?

A wave is a phenomenon that flows across a material medium without leaving any lasting mark.

Mechanical or electromagnetic waves can be used to classify them. Transverse and longitudinal are the two main forms.

A wave carries energy from one place to another. Mechanical waves carry energy through a material medium.

Hence the correct options for the fill-in-the-blanks are energy and a material medium.

To learn more about the wave refer to the link;

brainly.com/question/3004869

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