1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stepan [7]
3 years ago
13

Every day a certain amount of water evaporates from Earth’s oceans, lakes, and land surface and forms water vapor and clouds in

the atmosphere. Every day a certain amount of rain falls back to Earth. Make the reasonable assumption that, on average, the energy absorbed by the evaporation and lifting of the water is equal to the energy released by its condensation and falling back to earth. The evaporation of one mole of water requires approximately 40.6×103 J of heat. An equal amount of heat is released when a mole of water condenses.
QUESTION: The annual volume of rainfall on Earth is approximately 4.9×105 km3 and the average cloud altitude is 8.8 km above Earth’s surface. How much energy, in joules, is required every day to evaporate and lift the water?
Physics
1 answer:
pochemuha3 years ago
8 0

Answer:

0.0000000010 joules

Explanation:

Amount : 1540.3 nanojoules (nJ)

Equals : 0.0 joules (J)

You might be interested in
Giving quadrilateral a(2,-1 ) b ( 1,3) c(6,5) d(7,1) you want to prove that it is a parallelogram by showing opposite sides are
solong [7]

Answer:

Opposite sides are congruent (AB = DC).

Opposite angels are congruent (D = B).

Consecutive angles are supplementary (A + D = 180°).

If one angle is right, then all angles are right.

The diagonals of a parallelogram bisect each other.

Each diagonal of a parallelogram separates it into two congruent triangles.

Explanation: #if you need any queshtions answered within secs/mins hit me up and I gotchu.

8 0
3 years ago
PLEASE HELP!!!!!!!! More science questions coming soon!!
Naily [24]

Answer:

1.) A - 1,2,1,2

2.) Both forces can act without objects touching.

3.) C

Both can be modeled as waves having amplitude, frequency and wave lengths.

4.) A, B and C

Explanation:

1.) SO2 + ___ H2 ----------> _____ S + _______H2O

Looking at above equation, we have one molecule of surfur at the right hand side (RHS), and one molecule of surfur at the left hand side (LHS).

Two atoms of hydrogen and both RHS and LHS

But oxygen is not balanced. We have two atoms of oxygen at the RHS while having one at the LHS.

So let's make oxygen 2 atoms at LHS by adding 2 to water molecules and hydrogen molecules at the RHS.

SO2 + ___ 2H2 ----------> _____ S + _______2H2O

The correct answer is A - 1,2,1,2

2.) Both magnetic force and gravitational force obey inverse square law with distance. They are not directly proportional but inversely proportional to the square distance. B is the correct answer because Both forces can act without objects touching.

3.) Light waves are transverse waves which can travel through a vacuum without a medium while sound waves are longitudinal waves which cannot travel through vacuum without a medium. But both can be modeled as waves having amplitude, frequency and wave lengths.

4.) If an object is slowing down, due to conservation of energy,

The potential energy could be increasing, like a ball thrown into the air.

The kinetic energy could be lost to friction or air resistance.

The ball could be returning to its natural resting state.

7 0
3 years ago
a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the
ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

5 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
The attraction of liquid particles for a solid surface is due to ____.
White raven [17]
This attraction occurs from adhesion, also known as adsorption <span />
6 0
3 years ago
Read 2 more answers
Other questions:
  • What energy transformations from eating an apple to riding your bike?
    8·2 answers
  • Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele
    11·1 answer
  • When a plant is entering the calvin cycle of photosythesis______.
    7·2 answers
  • Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5^degree away
    13·1 answer
  • Which of these is important to gathering and interpreting scientific information?
    5·2 answers
  • Carlotta makes a graphic organizer about the steps of the Sun’s hydrogen fusion process. A venn diagram of 3 intersecting circle
    14·1 answer
  • Can someone in the fewest SIMPLE words possible, compare nuclear fusion and fission in terms of what happens in the reactions, t
    5·2 answers
  • 10. Ron’s father finally installed cruise control on their flying car. He was having trouble because he kept of forgetting that
    10·1 answer
  • Which is an example of a physical change?
    10·2 answers
  • A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!