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3 years ago

Extra glucose is _____.

Physics

2 answers:

Deffense [45]3 years ago

5 0

<em>2 weeks late :')</em>

<em>the person above me is right tho</em>

Murrr4er [49]3 years ago

3 0

Answer: stored in roots, stems, and leaves

Explanation:

As a result of the process of photosynthesis where carbon dioxide and water and used to synthesize glucose, the plant will find itself with extra glucose which is still needed but not at that point.

It will therefore store the glucose in roots, stems and leaves. It will however, convert them to starch first so that they are not affected by osmosis. Starch is not soluble in water which is why osmosis will not affect them and cause they to swell too much which would reduce the space the plant has.

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characterization of paraffin/ultrasonic-treated diatomite for use as phase change material in thermal energy storage of building

jarptica [38.1K]

The characterization of paraffin/ultrasonic-treated diatomite for use as phase change material in thermal energy storage of buildings and it provides the highest surface area with least to no structural degradation.

<h3>What are the characterization of paraffin/ultrasonic-treated diatomite?</h3>

Paraffin/ultrasonic-treated diatomite is used as a phase change material or PCM in thermal energy storage of buildings.

The diatomite is treated multiple times in ultrasonic sound for approximately sixty minutes or DA-60.

This is done to achieve the optimum condition to improve the surface area.

When the surface area reaches the highest, the structural degradation reaches minimum.

59°C is the melting point of the DA-60 composite phase change material. The latent heat of the same is 45.90 $Jg^-^1$ .

A stable PCM will have proper thermal reliability which is attained through thermal cycling test of 500 cycles.

To learn more about thermal energy, refer to:

brainly.com/question/7541718

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8 0

2 years ago

Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại v

tamaranim1 [39]

Sorry don’t understand your language.

6 0

3 years ago

A 750-kg car moving at 23 m/s brakes to a stop. Assume that all the kinetic energy is transformed into thermal energy. The brake

ahrayia [7]

Answer:

The temperature of the brakes is 29.38°C.

Explanation:

Given that,

Mass of car = 750 kg

Speed = 23 m/s

Mass of iron = 15 kg

We need to calculate the kinetic energy of car

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$K.E=\dfrac{1}{2}\times750\times23^2$

$K.E=198.375\ kJ$

We need to calculate the temperature of the brakes

Using formula of energy

$K.E=mc\Delta T$

Put the value into the formula

$198.375=15\times0.450\times\Delta T$

$\Delta T=\dfrac{198.375}{15\times0.450}$

$\Delta T=29.38^{\circ}C$

Hence, The temperature of the brakes is 29.38°C.

7 0

4 years ago

The emf of the battery is 1.5 V. In Nichrome there are 9 × 1028 mobile electrons per m3, and the mobility of mobile electrons is

kolezko [41]

Answer:

The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

Explanation:

Given;

emf of the battery, V = 1.5 V

electron density, = 9 × 10²⁸ mobile electrons per m³

mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)

length of thin wire, L = 6 cm = 0.06 m

cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²

The magnitude of the electric ﬁeld in the thin wire is given by;

E = V/L

E = (1.5) / (0.06)

E = 25 N/C

the number of electrons entering the thin wire every second is given by;

$e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s$

Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

5 0

3 years ago

A 0.40-kg cart with charge 4.0 x 10-5 C starts at rest on a horizontal frictionless surface 0.50 m from a fixed object with char

maw [93]

Answer:

26.82m/s

Explanation:

Given

Mass = m= 0.4kg

Initial Velocity = u = 0

Charge = 4.0E-5C

Distance= d = 0.5m

Object Charge = 2E-4C

First, we'll calculate the initial energy (E)

E = Potential Energy

PE = kQq / d

Where k = coulomb constant = 8.99E9Nm²/C²

Energy is then calculated by;

PE = 8.99E9 * 4E-5 * 2E-4 / 0.5

PE = 143.84J

Energy = Potential Energy = Kinetic Energy

K.E = ½mv² = 143.84J

½mv² = ½ * 0.40 * v² = 143.85

0.2v² = 143.85

v² = 143.85/0.2

v² = 719.25

v = √719.25

v = 26.81883666380777

v = 26.82m/s

Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge

4 0

4 years ago