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3 years ago

You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

Physics

2 answers:

enyata [817]3 years ago

6 0

Less than true weight

Darina [25.2K]3 years ago

5 0

Answer: A

Explanation: Your body gets pulled down, causing ur body to have more weight but not mass.

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Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17Â°cw with respect

mash [69]

The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.

<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.

The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.

<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

Learn more about intensity of polarized light here:

brainly.com/question/25402491

5 0

2 years ago

A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(

Flura [38]

Answer:

8.97 Watt

Explanation:

Resistance, R = 20 ohm

Inductance, L = 10 mH

V(t) = 20 Cos (1000 t + 45°)

Compare with the standard equation

V(t) = Vo Cos(ωt + Ф)

Ф = 45°

ω = 1000 rad/s

Vo = 20 V

Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm

impedance is Z.

$Z = \sqrt{R^{2}+X_{L}^{2}}$

$Z = \sqrt{20^{2}+10^{2}}$

Z = 22.36 ohm

$V_{rms}=\frac{V_{0}}{\sqrt{2}}$

$V_{rms}=\frac{20}{\sqrt{2}} = 14.144 V$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{14.144}{\sqrt{22.36}}=0.634 A$

Apparent power is given by

P = Vrms x Irms

P = 14.144 x 0.634

P = 8.97 Watt

6 0

4 years ago

A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?

vovikov84 [41]

We can solve the problem by using Newton's second law of motion:
$F=ma$
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
$a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2$

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>

5 0

3 years ago

An object moves from point A to point C along the rectangle shown in the figure below.

Naily [24]

Answer:

Hello friend where is the figure of the question

4 0

3 years ago

A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr

jasenka [17]

Answer: The original temperature was

$T_{1}=126.51K$