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An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.3

jekas [21]

Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0 N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C

4 0

3 years ago

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

What is the displacement of a spring if it has a spring constant of 10 N/m, and a force of 2.5 N is applied?

pochemuha

O.25 m is the displacement

5 0

3 years ago

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I need help on this please help me

daser333 [38]

A is growth!!!!! B is reproduction!!!

3 0

3 years ago

An object travels with a constant speed in a circular path. The net force on the object is

Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

$F=ma$

where m is the mass of the object.

In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

8 0

3 years ago