Which of the following expresses the answer to

Calcula la fuerza magnética de una carga que se desplaza hacia la derecha con una velocidad de 5 X 10 a la 6 metros sobre segund

Diano4ka-milaya [45]

Answer:

La fuerza magnética del electrón es 6.41x10⁻¹³ N.

Explanation:

Podemos encontrar la fuerza magnética (F) usando la ley de Lorentz:

$F = qv\times B$

En donde:

q: es la carga del electrón = 1.602x10⁻¹⁹ C

v: es la velocidad del electrón = 5x10⁶ m/s

B: es el campo magnético = 0.8 T

$F = qv\times B = 1.602 \cdot 10^{-19} C*5\cdot 10^{6} m/s*0.8 T = 6.41 \cdot 10^{-13} N$

Por lo tanto, la fuerza magnética del electrón es 6.41x10⁻¹³ N.

Espero que te sea de utilidad!

3 0

2 years ago

Energy conservation

balandron [24]

Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

Solution:

To solve this problem, we are going to employ the appropriate motion equation.

We must understand that this fall occurs in the presence of gravity;

V = U + 2gH

Where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height of the pool table

Since U is the unknown, let us make it the subject of the expression;

U = V - 2gH

U = 6 - (2 x 9.8 x 0.68) = 7.328m/s(deceleration)

7 0

3 years ago

What is difference between non uniform and uniform circular motion?

statuscvo [17]

Answer:

The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.

Explanation:

The speed of the object traveling in uniform motion is constant, the actual speed and the average speed of the moving body is same.

6 0

3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?

Shkiper50 [21]

If you wrap some of the wire around the nail in one direction and some of the wire in the other direction, the magnetic fields from the different sections fight each other and cancel out, reducing the strength of your magnet.

8 0

3 years ago