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3 years ago

What is the average distance from the Sun (in astronomical units) of a planet with an orbital period of 45.66 years?

1 answer:

6 0

If an object is orbiting the Sun with an orbital period of 15 years, what is its average distance from the Sun? a = (225)1/3 = 6.1 AU.
...
There is a simplified version of this law: P2 = a3 where:
The object must be orbiting the Sun.
P = period of the orbit in years.
a = average distance of the object from the Sun in AU.

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A lens with f = +11cm is paired with a lens with f = −25cm. What is the focal length of the combination?

Zepler [3.9K]

Answer:

19.642 cm

Explanation:

f₁ = Focal length of first lens = 11 cm

f₂ = Focal length of second lens = -25 cm

Combined focal length formula

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm$

Combined focal length is 19.642 cm

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3 years ago

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3 years ago

a vector is 253 m long and points in a 55.8 degree direction, what’s the y and x- component of the vector?

BartSMP [9]

Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

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3 years ago

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3 years ago

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

Yuri [45]

Answer:

$q = 2.067 \times 10^{-5}\ C$

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

$q =\dfrac{mg}{E}$

$q =\dfrac{0.00141 \times 9.81}{670}$

$q = 2.067 \times 10^{-5}\ C$

Charge of the particle is equal to $q = 2.067 \times 10^{-5}\ C$

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4 years ago