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4 years ago

What is the electric field at the position (x2,y2)=(−5.0cm , 0cm) in component form?

1 answer:

4 0

Complete Question: A +10 nC charge is located at the origin. What is the electric field at the position (x2,y2)= (-5.0cm, 0cm)? Write electric field vector in component form.

E2x, E2y=? N/C

Answer:

Ex = -3.6*10⁴ N/C Ey=0

Explanation:

As the charge producing the field is positive, the direction of the field, which is the one that would take a positive test charge located in the point of interest, will be away from this point, pointing to the left.

If we choose the positive direction to be to the right, the electric field component along the x-axis will be negative.

The magnitude of the field can be obtained applying the electric field definition, and the Coulomb's Law to the charge in the origin, as follows:

E = k*q/r² = 9*10⁹N*m²/C²*10⁻⁸C/(0.05)² m² = 3.6*10⁴ N/C

As the electric field follows the same line as the electric force, it has only component on the x axis, so:

Ex = -3.6*10⁴ N/C

Ey = 0

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Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin

Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

w₀B = ½ w₀

αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

t´ = t - t_1

et's write the equation of particle B

θ = θ₀ + w₀B t´ + ½ αB t´2

replace

θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0

3 years ago

When the pressure on a gas increases,what does the volume do?

kupik [55]

Answer:

As pressure goes up, volume goes down.

Explanation:

Pressure and volume of a gas are inversely proportional. This means that as pressure goes up, volume goes down. And as volume goes up, pressure goes down.

Cheers.

4 0

3 years ago

Read 2 more answers

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$