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Simora [160]
3 years ago
5

If two polarizing filters are held with their polarization axes at right angles to each other, the amount of light transmitted c

ompared to when their axes are parallel is
? a


zero.


? b


half as much.


? c


the same.


? d


twice as much.
Physics
1 answer:
Vinvika [58]3 years ago
3 0
Most likely answer is d. It would be half the size since the poliraztion is greater due to perpendicular bisect.
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How do physicists define velocity?
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Mars’s moon Phobos orbits the planet at a distance of 9380 km from its center, and it takes 7 hours and 39 minutes to complete o
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Answer: 0.107

Explanation:

We can solve this problem with Kepler's Third Law of Planetary motion:

T^{2}=4 \pi^{2} \frac{r^{3}}{G M_{MARS}} (1)

Where:

T=7 h 39 min is the orbital period of Phobos around Mars

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M_{MARS} is the mass of Mars

r=9380 km \frac{1000 m}{1 km}=9,380,000 m is the semimajor axis of the orbit Phobos describes around Mars (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

Well, firstly we have to convert the orbital period to seconds:

T=7 h 39 min=(7 h \frac{3600 s}{1 h}) + (39 min \frac{60 s}{1 min})=25200 s + 2340 s=27540 s

Now, we have to find M_{MARS} from (1):

M_{MARS}=\frac{4 \pi^{2} r^{3}}{G T^{2}} (2)

M_{MARS}=\frac{4 \pi^{2} (9,380,000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}) (27540 s)^{2}} (3)

M_{MARS}=6.436(10)^{23} kg (4) This is the mass of Mars

On the other hand, it is known the mass of the Earth is:

M_{EARTH}=5.972(10)^{24} kg (5)

Then, if we want to know the ratio of Mars’s mass to the mass of the earth, we have to divide M_{MARS} by M_{EARTH}:

\frac{M_{MARS}}{M_{EARTH}}=\frac{6.436(10)^{23} kg}{5.972(10)^{24} kg}

Finally:

\frac{M_{MARS}}{M_{EARTH}}=0.107

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A car starts from rest and accelerates at 2.5m/s^2 for 2s. What is the final velocity of the car
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Answer:

5m/s ez

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