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inn [45]
3 years ago
10

9) An object with a height of 18 cm is placed in front of a converging lens. The image has a

Physics
2 answers:
vovikov84 [41]3 years ago
6 0

Answer:

Explanation:

a) Magnification = image height / object height = -9 / 18 = -0.5

b) Magnification = - image distance / object distance = -0.5

so image distance = 0.5 object distance

1/focal length = 1/image distance + 1/object distance

1/6 = 1/(0.5 object distance) + 1/object distance

object distance = 18.0 cm

c) Image appears behind the lens.

lorasvet [3.4K]3 years ago
5 0

Answer:

Explanation:

mag = ht i/ht o = -9/18 = -1/2

mag = -1/2 = -di/do

do = 2di

put in eqn

1/f = 1/di + 1/do

1/6 = 1/di + 1/2di

2di = 18

do = 18cm

image is behind lens

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All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?
irinina [24]

Answer:

See the answer below

Explanation:

The optimal conditions for high biodiversity seem to be a <u>warm temperature</u> and <u>wet climates</u>.

<em>The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 </em>^oC<em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>

Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 ^oC and wet environment with annual precipitation of not less than 262 cm.

The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.

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3 years ago
If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

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Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

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Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

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