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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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Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred

madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas P V = m R T

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

$P_2=\dfrac{T_2}{T_1}\times P_1$

$P_2=\dfrac{600}{300}\times 2$

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

$m=\dfrac{PV}{RT}$

$m=\dfrac{200\times 3}{0.287\times 300}\ kg$

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0 Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

$S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}$

$S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}$

S₂ - S₁ = 3.42 KJ/K

5 0

2 years ago

A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator

cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance

= mass * gravity * distance

= 1000 * 9.81 * 10

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J

Work done by cable = PE +KE

108,100 J = KE + 98,100 J

KE = 10,000 J

= 10kJ

7 0

3 years ago

Read 2 more answers

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

Oxana [17]

Answer:

A) $I_{total}$ = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = $I_{cm}$ + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

$I_{total}$ = $I_{body}$ + 2 $I_{arm}$

$I_{body}$ = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

$I_{arm}$ = M D²

Let's write the equation

$I_{total}$ = ½ M R² + 2 (m D²)

Let's calculate

$I_{total}$ = ½ 56 0.20² + 2 4 0.20²

$I_{total}$ = 1.12 + 0.32

$I_{total}$ = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

2 years ago

The number of waves that pass a given point in a second is___.

Kisachek [45]

Answer:

Frequency

Explanation:

The frequency ( ) of a wave is the number of waves passing a point in a certain time.

5 0

2 years ago