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2 years ago

7

An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 50 km 60.0° west of north, and t

he second is 100 km 30.0° east of north. What is the total displacement of the airplane?

1 answer:

Rashid [163]2 years ago

8 0

Answer:

LOL

Explanation:

IMAGINE POSTING UR CLASSWORK LOLL

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Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge

butalik [34]

Answer:

Electric charge in the earth will be $Q=5.231\times 10^5C$

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}$ . here K is coulomb's constant

So $116=\frac{9\times 10^9\times Q}{(6371000)^2}$

$Q=5.231\times 10^5C$

So electric charge in the earth will be $Q=5.231\times 10^5C$

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3 years ago

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bonufazy [111]

Answer:

The intensity of laser 2 is 4 times of the intensity of laser 1.

Explanation:

The intensity in terms of electric field is given by :

$U=\dfrac{1}{2}\epsilon_o E^2$

E is electric field

It means, $U\propto E^2$

In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.

Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

$\dfrac{U}{U'}=(\dfrac{E}{E'})^2$

We have,

E'=2E

So,

$\dfrac{U}{U'}=\dfrac{E^2}{(2E)^2}\\\\\dfrac{U}{U'}=\dfrac{E^2}{4E^2}\\\\\dfrac{U}{U'}=\dfrac{1}{4}\\\\U'=4\times U$

So, the intensity of laser 2 is 4 times of the intensity of laser 1.

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D subtract the atomic number from the atomic mass

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Answer:

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