All categories

2 years ago

These materials would result in horizontally polarized light.

1 answer:

7 0

The correct answers among all the other choices are D.) reflection from wet asphalt and E.) refraction from a water surface. These materials would result in horizontally polarized light. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

You might be interested in

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

3 years ago

A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

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

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is:

$\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}$

$\alpha = 36.469\,\frac{rad}{s^{2}}$

$I = \frac{\tau}{\alpha}$

$I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }$

$I = 0.0823\,N\cdot m\cdot s^{2}$

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0

2 years ago

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper e

Mashcka [7]

Answer:

Explanation:

When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.

This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,

T cos θ will provide the centripetal force . So

Tcosθ = mw²R

Tsinθ component will balance the weight .

Tsinθ = mg

Dividing the two equation

Tanθ = $\frac{g}{\omega^2R}$