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lakkis [162]
3 years ago
9

Clouds dump around 100 billion gallons of water on rainforests each year. How much rain is evaporated from the rivers, lakes and

surface of rainforests each year?
A) around 100 billion gallons
B) unable to tell from the information given
C) significantly less than 100 billion gallons
D) significantly more than 100 billion gallons
Physics
2 answers:
MakcuM [25]3 years ago
5 0

 

A)  around 100 billion gallons

posledela3 years ago
3 0
The correct answer is this one: "D) significantly more than 100 billion gallons ." Clouds dump around 100 billion gallons of water on rainforests each year. The amount of  rain is evaporated from the rivers, lakes and surface of rainforests each year is significantly more than 100 billion gallons<span> </span>
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An astronaut has a mass of 50.0 kg on earth. what is her mass on the moon, where gravity is 1 6 that on earth?
never [62]

I believe the correct gravity on the moon is 1/6 of Earth. Take note there is a difference between 1 6 and 1/6.

HOWEVER, we should realize that the trick here is that the question asks about the MASS of the astronaut and not his weight. Mass is an inherent property of an object, it is unaffected by external factors such as gravity. What will change as the astronaut moves from Earth to the moon is his weight, which has the formula: weight = mass times gravity.

<span>Therefore if he has a mass of 50 kg on Earth, then he will also have a mass of 50 kg on moon.</span>

6 0
4 years ago
Drag each label to the correct location. Sort the sentences based on whether they describe the properties of a heterogeneous or
Lisa [10]

Answer:

ijpferiukjlwbl aojh oljn,

Explanation:

5 0
3 years ago
Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
4 years ago
The three branches of science earth and physicals
seropon [69]
<span>Science is always defined as the body of knowledge and linked to something related to the physical world, phenomena and lives of everything that exist on Earth. Science is categorized in various ways.  Pure, practical or theoretical, is the largest distinction of  science or whether science is applied.  Generally, pure science is subdivided into Physical sciences, Earth sciences, and Life sciences.  Physical sciences deal with energy and matter, Earth sciences explain Earth’s phenomena, to the solar system where it belongs and its atmosphere. And finally, life sciences describe internal processes of living organisms, its relationship to each other and to its environment.</span>


6 0
3 years ago
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