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Elina [12.6K]
2 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]2 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]2 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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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
djyliett [7]
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
A: mgh_1+\frac{mv_1^2}{2}
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
W_f=F_f\cdot L
So, we know that the car is approaching the point B with the following amount of energy:
mgh_1+\frac{mv_1^2}{2}- F_fL
The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
B: mgh_2+\frac{mv_2}{2}
As said before this energy must be the same as the energy of a car approaching the loop:
mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}

4 0
2 years ago
Read 2 more answers
A solid cube of edge length r, a solid sphere of radius r, and a solid hemisphere of radius r, all made of the same material, ar
juin [17]

Answer:

Explanation:

The rate at which heat will be radiated is given by the expression

E = e Aσ ( T⁴ - T₀⁴ )

E is heat radiated , e is emissivity , A is area of surface , σ  is stephan's constant T is temperature of the object and T₀ is temperature of the surrounding .

For all the objects given , e , σ T and T₀ are same so E will solely dependent on area of the surface

surface area of cube= 6 r² ,

surface area of sphere = 4 π r²

= 12.56 r²

hemisphere = 2 π r²

= 6.28 r²

12.56 r² >6.28 r² > 6 r²

heat radiated by sphere > heat radiated by hemisphere > heat radiated by cube .

8 0
2 years ago
Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mas
Elan Coil [88]
<span>Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>
7 0
2 years ago
Physics professor Antonia Moreno is pushed up a ramp inclined upward at an angle 31.0 ∘ above the horizontal as she sits in her
DerKrebs [107]

Answer:

The answer is below

Explanation:

Given that:

mass (m) = 86 kg, distance (L) = 2.75 m, θ = 31°, force (F) = 595 N, initial velocity (v_i) = 2.4 m/s, g = acceleration due to gravity = 9.8 m/s²

The net work can be gotten from the equation:

W_{net}=Fcos(\theta)L-mgsin(\theta)L\\\\W_{net}=595*cos(31)*2.75-[86*9.81*sin(31)*2.75)\\\\W_{net}=1402.54-1194.92\\\\W_{net}=207.62

From the work-energy theorem equation, we can get her speed at the top of the ramp (v_f)

Hence:

W_{net}=change\ in\ kinetic\ energy\\\\W_{net}=\frac{1}{2}m(v_f^2-v_i^2 )\\\\2W_{net}=m(v_f^2-v_i^2 )\\\\v_i=\sqrt{ v_i^2+\frac{2W_{net}}{m}} \\\\v_f=\sqrt{ 2.4^2+\frac{2*207.62}{86}}}\\\\v_f=3.25\ m/s

8 0
2 years ago
What voltage is required to move 6A through 20?<br><br>​
Marina CMI [18]

Answer:

120V

Explanation:

Given parameters:

Current  = 6A

Resistance  = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

           V = IR

Where V is the voltage

            I is the current

            R is the resistance

Now, insert the parameters and solve;

    V  = 6 x 20  = 120V

7 0
2 years ago
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