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3 years ago

Which form of energy does a plant store when light is transformed during photosynthesis into sugars?

Physics

1 answer:

AlexFokin [52]3 years ago

4 0

Answer:

C. Chemical energy

Explanation:

The different types of energy listed in this question are:

A) Electrical energy: it is the energy related to the flow of electrical charges (current)

B) Nuclear energy: it is the energy contained in the nuclei of the atoms, and released during nuclear reactions

C) Chemical energy: it is the energy contained in the bonds between molecules, and it is released during chemical reactions

D) Thermal energy: it is the energy associated with the motion of molecules

Photosynthesis is the process that plants use to transform the light energy (coming from the Sun) into chemical energy. In fact, in this process, plants absorb CO2 (from the atmosphere) and solar energy (from the light), and they convert them into sugars (glucose) and oxygen. Therefore, the initial energy coming as solar energy is converted into energy of the bonds of the molecules of glucose (so, chemical energy).

So, the correct option is C).

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3 years ago

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Answer:

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Explanation: When a liquid changes into a gas vaporization has occurred. The process can either occur due to boiling or evaporation. Boiling occurs when the vapor pressure of the liquid is raised (by heating) to the point where it is equal to the atmospheric pressure.

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6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the

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2 years ago

A 580-turn solenoid is 18 cm long. The current in it is 36 A. A straight wire cuts through the center of the solenoid, along a 2

Karolina [17]

Answer:

F = 0.078N

Explanation:

In order to calculate the magnitude of the force on the wire you first calculate the magnitude of the magnetic field generated by the solenoid, by using the following formula:

$B=\frac{\mu_oNi}{L}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 580

i: current in the solenoid = 36A

L: length of the solenoid = 18cm = 0.18m

You replace the values of all parameters in the equation (1):

$B=\frac{(4\pi*10^{-7}T/A)(580)(36A)}{0.18m}=0.145T$

Next, you calculate the force exerted on the wire, by using the following formula:

$F=iLBsin\theta$ (2)

i: current in the wire = 27A

L: length of the wire that perceives the magnetic field (the same as the radius of the solenoid) = 2.0 cm = 0.02m

θ: angle between wire and the direction of B

B: magneitc field in the solenoid = 0.145T

The direction of the wire are perpendicular to the direction of the magnetic field, hence, the angle is 90°.

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

$F=(27A)(0.02m)(0.145T)sin90\°=0.078N$

The magnitude of the force on the wire is 0.078N

8 0

3 years ago

Read 2 more answers

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago