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3 years ago

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A jet plane is moving at a constant velocity on a flat surface. Which forces act against the forward motion of the plane? A. Gra

vity and engine thrust B. Engine thrust and friction C. Friction and air resistance D. Air resistance and gravity

1 answer:

Leni [432]3 years ago

5 0

Answer:

C Friction and air resistance

Explanation:

The force of friction acts in a direction opposite to the direction of motion of an object in contact with a surface. The amount of friction is proportional to the moving object's weight and a coefficient of friction (which is dependent on the object-surface contact material).

Air resistance, similar to surface-induced friction, acts as a deceleration force. Its magnitude depends on the object's speed and surface area.

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago

__________ force is a fictitious force that some people believe causes you to feel as if you are being pushed outward from the c

jok3333 [9.3K]

The complete sentence is:
<span>Centrifugal force is a fictitious force that some people believe causes you to feel as if you are being pushed outward from the center of a circle while traveling in uniform circular motion.

In fact, centrifugal force is an inertial force: it is not a real force, but it is due to the fact that the reference frame is rotating. The real force in a uniform circular motion is the centripetal force, which pushes towards the centre of the circle, and keeps the object in circular motion.</span>

5 0

3 years ago

Read 2 more answers

The moon Umbriel orbits Uranus (mass = 8.68 x 10^25 kg) at a distance of 2.66 x 10^8 m. What is Umbriel's orbital period (in hou

Whitepunk [10]

Answer:

T = 99.51 hour

Explanation:

Mass of Uranus, $M=8.68\times 10^{25}\ kg$

The moon Umbriel orbits Uranus at a distance of $2.66\times 10^8\ m$

We need to find Umbriel's orbital period. Let it is T. Using Kepler's third law of motion to find it.

$T^2\propto r^3\\\\T^2=\dfrac{4\pi^2r^3}{GM}\\\\T^2=\dfrac{4\pi^2\times (2.66\times 10^8)^3}{6.67\times 10^{-11}\times 8.68\times 10^{25}}\\\\T=358244.51\ s$

As 1 hour = 3600 s

358244.51 s = 99.51 hour

Hence, Umbriel's orbital period is 99.51 hour.

7 0

3 years ago

Which of these waves cannot travel in a vacuum?

gayaneshka [121]

Light. The black hole is a vacuum, and not even light can escape it's suction

4 0

3 years ago

Read 2 more answers

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago