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Kipish [7]
3 years ago
7

Create a story of a “day in the life of a water drop”. Your story should be creative and include all 6 steps in the water cycle.

You can share your story through, writing (1 paragraph per part of the cycle), drawing, animation, or a multimedia presentation.
Physics
2 answers:
klasskru [66]3 years ago
8 0

Answer: Once upon a time the clouds were coming together for a family meeting. One uncle cloudy got mad at his brother and started a rumble. As he got angrier and angrier the clouds started to shake. Later down below in humans vile a storm started to form. And that was the day a rain drop was born. :)

Explanation:

pogonyaev3 years ago
5 0

Answer:

Explanation:

Ok so first: the evaporation part, the sun starts to get warmer I ( the water droplet) rises up to the sky to start my evaporation cycle

Condensation: part: when I am in the air I change into a gas and then I change back into a liquid and gather my friends and make a cloud

Precipitation: as it gets to crowded, we can’t hold it anymore, when I cool down I like to sky dive with my cousins, snow, rain, sleet, hail which is called precipitation.

Then finally we land on the ground, we run down hills, and run into lakes surface runoff happens when there’s too many of us so some of us can’t be rain. Infiltration: when some of us soak into the ground cause we can’t make it into the streams and oceans. Ok I can’t help much more cause I’m super busy but if you need more help just message me and I can help thx ! Hope I helped Atleast a bit for you to understand more

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Write down the principals of them lever in points .​
Juli2301 [7.4K]

The major principal of leaver is

load × load distance = effort × effort distance

where,

effort dis= distance between effort and fulcrum

load distance = distance between load and fulcrum......

6 0
2 years ago
The y component of a vector R of magnitude k = Bcm shown in the figure below is Ky = +6 cm. What is the direction of this vector
WARRIOR [948]

Given

The y-component of vector K is

K_y=6\text{ cm}

The magnitude of vector K is , K=8 cm

To find

The angle

\theta

Explanation

Resolving K along its y-component we have,

\begin{gathered} K_y=Ksin\theta \\ \Rightarrow6=8sin\theta \\ \Rightarrow sin\theta=\frac{3}{4} \\ \Rightarrow\theta=48.59\text{ }^o \end{gathered}

Conclusion

The angle made with the x-axis is

48.59^o

6 0
1 year ago
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3 years ago
Read 2 more answers
Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a
Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }

Sum of the potential at point p is

V = Va + Vb

that is

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]

the expression below can be written as the equivalent

\frac{1}{(a^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }

likewise,

\frac{1}{(b^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }

hence,

V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

by reciprocal rule

1/a² = a⁻²

V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]

by binomial expansion of fractional powers

where (1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...

if we expand the expression we have the equivalent as shown

{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )

also,

{(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )

the above equation becomes

V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]

V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )

V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )

Answer

V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }

OR

V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }

8 0
3 years ago
Determine the density of a rectangular piece of concrete that measures 3.7 cm by 2.1 cm by 5.8 cm and has a mass of 43.8 grams.
Novay_Z [31]
It is customary to work in SI units.

Calculate the volume of the concrete.
V = 3.7*2.1*5.8 cm³ = 45.066 cm³ = 45.066 x 10 ⁻⁶ m³

The mass is  43.8 g = 43.8 x 10⁻³ kg

The density is mass/volume.
Density = (43.8 x 10⁻³ kg)/(45.066 x 10⁻⁶ m³) = 971.9 kg/m³

Answer: 971.9 kg/m³
5 0
3 years ago
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