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Elina [12.6K]
3 years ago
12

The descriptions below explain two ways that water is used by plants on a sunny day. I. In a process called transpiration, some

liquid water in leaves changes to water vapor. The water vapor is released into the air through tiny pores in the leaves. This allows more liquid water from the soil to be pulled up the roots and stem to replace water lost from the leaves. II. Plants use some of this water in leaves in a process called photosynthesis. During photosynthesis, water and carbon dioxide break apart and recombine to form two new substances, oxygen and glucose. Based on the above description of transpiration and photosynthesis, which type of change happens to water during each process?
Physics
2 answers:
Anit [1.1K]3 years ago
6 0

Answer:

A- In transpiration, because some of its properties change, water undergoes a physical change but keeps its identity. In photosynthesis, because its identity changes, water undergoes a chemical change.

Explanation:

Dont Ask I Know Means i KNOW! Im Serious

grigory [225]3 years ago
4 1
In photosynthesis, the water is being used to create food for the plant (Glucose). In transpiration the water is going from a liquid to a gas that's being released.
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400 m/s

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80 x 5 = 400

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A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric
Tcecarenko [31]

Answer:

a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  \left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

       

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

           y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

           y = 1 - 2.04 10³

           y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

          z = 0 + 0 + ½ 5.1 10² 0.02²

          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + a_{y} t

         v_{y} = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

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