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DochEvi [55]
3 years ago
15

Please explain how I would do this!!!

Physics
1 answer:
nevsk [136]3 years ago
6 0

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

Explanation:

NMNMBNMNMN

You might be interested in
Why was Democritus considered a philosopher and not a scientist?
Alika [10]

Answer:

Democritus had no scientific instruments to extend the reach of his senses, so all of his experiments were just 'mind experiments', but because Democritus was a philosopher, he thought more into depth about why we humans are alive which led to the atomic theory.

Explanation:

5 0
3 years ago
What demonstrates the flow of energy and materials through
Tcecarenko [31]
A. Food chain

A food web also shows the flow of energy and materials through an ecosystem but in a complex web, not a single chain. A biogeochemical cycle does not deal with the flow of energy/materials through an ecosystem at all.
3 0
1 year ago
A 0.43 m long and 0.43 m wide loop is moved at a constant velocity of 0.15 m/s into a perpendicular constant magnetic field of 0
olya-2409 [2.1K]

Answer:

The magnitude of the induced voltage in the loop is 20 mV.

Explanation:

given;

length of loop, L = 0.43 m

width of loop,w = 0.43 m

velocity of moved loop, v = 0.15m/s

magnetic field strength,B = 0.31 T

To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;

magnitude induced E.M.F = BLv

magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV

Therefore, the magnitude of the induced voltage in the loop is 20 mV.

7 0
3 years ago
The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi
vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

Q=cm\Delta T (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and \Delta T the change in temperature of the sample. So, if we solve (1) for

Sample A:

\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}

\Delta T=-\frac{Q}{16744}

Sample B:

\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}

\Delta T=-\frac{Q}{5400}

Sample C:

\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}

\Delta T=-\frac{Q}{9450}

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression -\frac{Q}{cm} that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of \Delta T implies the temperature is dropping)

8 0
3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
3 years ago
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