3. Do you have an issue with the NSA monitoring citizens’ phone calls, text messages, emails, etc.?

Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of

steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$

Where,

$I_0$ = Initial Intensity

I = Final Intensity after pass through the polarizer

$\theta$ = Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

$\frac{I}{I_0} = \frac{1}{2}$

Using the Malus Law we have,

$I = I_0 cos^2\theta$

$cos^2\theta = \frac{I}{I_0}$

$cos^2\theta = \frac{1}{2}$

$\theta = cos^{-1}(\frac{1}{2})^2$

$\theta = 75.52\°$

Angle with respect to maximum is $90-75.52 = 14.48\°$

8 0

2 years ago

The ability to make things happen is also called _____.

DerKrebs [107]

Force bc it says the ability to make stuff happen

3 0

2 years ago

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

2 years ago

A 200g(m2) mass is pulling the cart (m1) across the table as shown below. If it covers 1.12m in 4.8 seconds starting from rest,

Vsevolod [243]

Answer: 20.21 kg

Explanation:

The mass hanging from the pulley is pulling the cart. The force by which cart is being pulled is equal to weight of the hanging mass.

$F=m_2g$

⇒F=0.200 kg ×9.8 m/s²=1.96 N

Since there is no other force acting on the cart and there is no friction, so this force will pull the cart.

The Cart covers 1.12 m distance in 4.8 s.

From the equation of motion,

s = u t + 0.5 × a t²

We will find the acceleration of the cart,

Distance covered, s = 1.12 m

Initial velocity, u = 0 ( staring from rest)

Time taken, t = 4.8 s

⇒1.12 m = 0 + 0.5 × a × (4.8 s)²

⇒a = 0.097 m/s²

Now the force which causes this acceleration is:

F = ma

where m is the mass of the cart and a is the acceleration of the cart

⇒1.96 N = m×0.097 m/s²

⇒m = 20.21 kg

Hence, the mass of the cart is 20.21 kg.

4 0

2 years ago

The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin

leva [86]

<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>

6 0

3 years ago