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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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A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0
1 year ago
A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc
Tasya [4]

Answer:

17 m/s south

Explanation:

m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 = Velocity of dog = 1 m/s

u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The velocity of the skateboard will be 17 m/s south as the north is taken as positive

3 0
2 years ago
An electron moves 4.5 m in the direction of an electric field of strength 325 N/C. What is the change in electrical potential en
Vinvika [58]
Calculate q* E * d 
<span>Put q = 1.6 x 10^-19 </span>
<span>E = 325 </span>
<span>d = 4.5 


I hope this helps!</span>
6 0
2 years ago
A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*
4 0
2 years ago
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