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3 years ago

Help plz I’ll mark brainliest

Physics

2 answers:

slavikrds [6]3 years ago

3 0

Thinner at edges and its thick in the middle

Nat2105 [25]3 years ago

3 0

A Piece Of Glass Which Is Thinner In The Middle Than On The Edges.
Hope This Helps You.

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Is this correct and plz j answer the right one I can’t have an explanation rn I’m in a rush thx so much

vova2212 [387]

I pretty sure it’s C

6 0

3 years ago

What is the difference between mechanical waves and electromagnetic waves?

Elina [12.6K]

Answer:

Explanation: Mechanical waves need a medium to travel.

3 0

3 years ago

A point charge of 3 µC is located at x = -3.0 cm, and a second point charge of -10 µC is located at x = +4.0 cm. Where should a

Nataly [62]

Answer:

The charge q₃ must be placed at X = +2.5 cm

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

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)

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

Equivalences

1µC= 10⁻6 C

1cm= 10⁻² m

Data

k = 8.99*10⁹ N×m²/C²

q₁ =+3 µC =3*10⁻⁶ C

q₂ = -10 µC =-10*10⁻⁶ C

q₃= +6µC =+6*10⁻⁶ C

d₁ = 3cm =3×10⁻² m

d₂ = 4cm = 4×10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

E₁:Field at point P due to charge q₁. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

E₂: Field at point P due to charge q₂. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

Problem development

E₃: Field at point P due to charge q₃. As the charge q₃ is positive, the field leaves the charge.

The direction of E₃ must be (- x) so that the electric field can be equal to zero at point P since E₁ and E₂ are positive, then, q₃must be located to the right of point P.

We make the algebraic sum of fields at point P due to the charges q1, q2, and q3:

E₁+E₂-E₃=0

$\frac{k*q_{1} }{d_{1}^{2} } +\frac{k*q_{2} }{d_{2}^{2} } -\frac{k*q_{3} }{d_{3}^{2} } =0$

We eliminate k

$\frac{q_{1} }{d_{1} ^{2} } +\frac{q_{2} }{d_{2} ^{2} }+\frac{q_{3} }{d_{3} ^{2} }=0$

We replace data

$\frac{3*10^{-6} }{(3*10^{-2})^{2} } +\frac{10*10^{-6} }{(4*10^{-2})^{2} } +\frac{6*10^{-6} }{d_{3} ^{2} } =0$

we eliminate 10⁻⁶

$\frac{3}{9*10^{-4} } +\frac{10}{16*10^{-4} } =\frac{6}{d_{3}^{2} }$

$(\frac{1}{10^{-4} }) *(\frac{1}{3} +\frac{5}{8}) =\frac{6}{d_{3}^{2} }$

$\frac{23*10^{4} }{24} =\frac{6}{d_{3} ^{2} }$

$d_{3} =\sqrt{\frac{6*24}{23*10^{4} } }$

$d_{3} =2.5*10^{-2} m\\d_{3} =2.5 cm$

The charge q₃ must be placed at X = +2.5 cm

6 0

3 years ago

Una persona sale de su casa y camina en linea recta 5m hacia la derecha, se para en una farola y gira 90 grados hacia la derecha

max2010maxim [7]

Answer:

El espacio recorrido es igual a 25 [m]

Explanation:

Supongamos que dicha persona se encuentra en el origen de un sistema de coordenadas cartesianas es decir en el punto (0,0). De tal manera que cuando se dirige hacia la derecha se mueve 5 metros, es decir ahora estará en el punto (5,0). Cuando gira hacia la derecha caminando en linea recta 20 metros, se moverá hacia el sur es decir estará en el punto (5,-20).

El total de espacio recorrido sera la suma de la distancia en cada uno de los desplazamientos 5 + 20 = 25 [m]

7 0

3 years ago

In comparing the gravitational force and the electrical force between two protons, the electrical force is _____ relative to the

EastWind [94]

The electrical force is greater.
I'm not going to look it up and work it out right now,
but it seems to me that the electrical force is bigger
by a factor of something like 10⁴⁰ . That's a lot !

8 0

3 years ago