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Svetlanka [38]
1 year ago
14

If I0 is the intensity of the unpolarized light incident on the first polarizer, and I1 and I2 denote the intensity of the light

after the first and second polarizers, respectively, which of the following quantities are known?
a) I1
b) theta1
c) I0
d) I2
e) theta2
Physics
1 answer:
e-lub [12.9K]1 year ago
7 0
E or C 10 hope this helps
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José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


8 0
1 year ago
What is the formula that can be used to find velocity if kinetic energy and mass are known?
viva [34]
The formula for kinetic energy is \frac{1}{2}m\Delta v^2. Thus, the equation for velocity is v=   \sqrt{ \frac{2TotalKineticEnergy}{m} }. 
6 0
1 year ago
Read 2 more answers
A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con
Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

8 0
2 years ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
1 year ago
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

   \frac{T}{r^{\frac{3}{2} } } = constant

For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
2 years ago
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