What is special about decomposers? *

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

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

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm

faltersainse [42]

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

$6=\frac{6}{365.25\times 24\times 60\times 60}$

So in one second it will move

$\frac{6}{365.25\times 24\times 60\times 60}$

In 71 seconds

$71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm$

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

$6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My$

Speed of the tectonic plate is 37.278 mi/My

7 0

3 years ago

¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula

Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

$F = |q|vBsin(\theta)$ (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

$v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s$

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.

Espero que te sea de utilidad!

7 0

3 years ago

What damage a tornado can cause

Ksivusya [100]

Alrighty, so, houses can be destroyed, people can be killed, animals can be killed, habitats can be destroyed, and trees can be uprooted. Tornadoes can create massive damage.

Hope this helps! (:

5 0

3 years ago

Describe what happens to water vapor when thermal energy is removed from it?

Marysya12 [62]

Explanation:

so sorry

don't know but please mark me as brainliest please

7 0

3 years ago