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Andrei [34K]
3 years ago
9

3. Describe the flow of one molecule of water through the water cycle, beginning in the ocean.

Physics
2 answers:
lubasha [3.4K]3 years ago
7 0

Answer:

Water travels in a continuous path through Earth called the water cycle. The journey starts in the Earth's oceans. As the Sun heats the ocean, the water evaporates into a gas. The gas meets cooler air, and the gas changes into droplets of tiny water molecules. This process is called condensation. Little droplets of water fall back down to Earth again as precipitation. The precipitation fills up the rivers and various other things and the cycle begins all over again.

Explanation:

scZoUnD [109]3 years ago
6 0

molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.

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3 years ago
A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T
maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: E_k=\frac{1}{2}mv^2

2. gravitational potential energy of mass m, grav. acc. g and height h: E_g=mgh

3. potential energy in a spring with spring constant k and displacement from equilibrium x: E_s=\frac{1}{2}kx^2

Calculating x:

\frac{1}{2}mv_a^2=\frac{1}{2}kx^2

x=\sqrt{\frac{m}{k}}v_a

Calculating the speed:

\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}

h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R

\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R

Solving for v_b:

v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}

7 0
3 years ago
What are (a) the x component and (b) the y component of a vector in the xy plane if its direction is 259° counterclockwise from
garik1379 [7]

Answer:

|Ax| =1.03 m  (directed towards negative x-axis)

|Ay|= 5.30 (directed towards negative y-axis)

Explanation:

Let A is a vector = 5.4 m

θ = 259°

to Find Ax, Ay

Sol:

according the condition it lies in 3rd quadrant

we know that Horizontal Component Ax = A cos θ

Ax = 5.4 Cos 259°

Ax = - 1.03 m

|Ax| =1.03 m  (directed towards negative x-axis)

Now Ay = A sin θ

Ay = 5.4 Sin 259°

Ay = -5.30

|Ay|= 5.30 (directed towards negative y-axis)

3 0
3 years ago
Read 2 more answers
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