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4 years ago

What part of the plant cell is responsible for making its food?

Physics

1 answer:

ArbitrLikvidat [17]4 years ago

4 0

It is a part of the sugary liquid in the plant that uses photosynthesis to produce.

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Two flat glass plates are separated at one end by a wire of diameter 0.20 mm; at the other end they touch. Thus, the air gap bet

stepan [7]

Answer:

668 bright fringes

Explanation:

t = Thickness = 0.2 mm

$\lambda$ = Wavelength = 600 nm

m = Number of fringes

We have the fringe width relation

$2t=\left(m+\frac{1}{2}\right)\lambda\\\Rightarrow m=\dfrac{2t}{\lambda}-\dfrac{1}{2}\\\Rightarrow m=\dfrac{2\times 0.2\times 10^{-3}}{600\times 10^{-9}}-\dfrac{1}{2}\\\Rightarrow m=666.166\approx 667$

So, total number of fringes will be including m = 0 is $1+667=668$

8 0

4 years ago

Pls help i have test

vova2212 [387]

Answer:

A 267 N forward

Explanation:

hope this helps

6 0

3 years ago

Read 2 more answers

4 examplels of chemical change?

kiruha [24]

Answer:

Burning of paper and log of wood.

Digestion of food.

Boiling an egg.

Chemical battery usage.

Electroplating a metal.

Baking a cake.

Milk going sour.

i did some extra for you

6 0

3 years ago

Read 2 more answers

A conducting rod (length = 2.0 m) spins at a constant rate of 2.0 revolutions per second about an axis that is perpendicular to

vladimir2022 [97]

Answer:

0.050V

Explanation:

To solve this problem it is necessary to use the concepts related to the potential between two objects that have a magnetic field, this concept is represented in the equation.

$\int dV = \int_{0}^{l/2} Bv (dl)$

Where,

v= tangencial velocity

B = Magnetic Field

We know for definition that,

$v= l\omega$

Where,

L = length

$\omega =$ Angular velocity

We can replace this values in our first equation then,

$\int dV = \int_{0}^{l/2} B (l\omega) (dl)$

Integrating we have,

$V = \frac{1}{8} Bl^2 \omega$

Replacing the values,

$V= \frac{1}{8} (8*10^{-3})(12.56)(4)$

$V = 0.050V$

Therefore the potential difference between the center of the rod and the other rod is 0.050V

7 0

3 years ago

A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the

Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

τ = 20µ x 1.5k = 30 ms

v = v₀e^(t/τ)

0.5 = 1.5e^(t/30ms)

e^(t/30ms) = 10/3

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0

4 years ago