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4 years ago

Green plants need light in order to survive. Structures in the leaves absorb light, which in turn, helps plants make

Physics

1 answer:

Nat2105 [25]4 years ago

3 0

Answer:

Explanation:

As shown in detail in the absorption spectra, chlorophyll absorbs light in the red (long wavelength) and the blue (short wavelength) regions of the visible light spectrum. Green light is not absorbed but reflected, making the plant appear green. Chlorophyll is found in the chloroplasts of plants.

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On March 21, a stick casts the following shadow. What is the most likely time of day?

jenyasd209 [6]

Answer:

9:00 AM

Explanation:

I took the test and that was the answer

5 0

4 years ago

Read 2 more answers

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

mina [271]

Answer:

(a) v = 463.97 m/s, $a_{c} = 0.0337\ m/s^{2}$

(b) v' = 196.02 m/s, $a'_{c} = 0.0143\ m/s^{2}$

Solution:

As per the question:

Time period of the rotation of earth, T = 24 h = $24\times 3600 = 86400\ s$

Radius of the earth, R = $6.38\times 10^{6}\ m$

Angle, $\theta = 65.0^{\circ}$

Now,

Angular velocity is given by:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{86400} = 7.27\times 10^{- 5} \rad$

(a) To calculate the speed and the centripetal acceleration at the equator:

Linear velocity or the speed, v = $R\omega$

$v = 6.38\times 10^{6}\times 7.27\times 10^{- 5} = 463.967\ m/s$

Centripetal Acceleration, $a_{c} = \frac{v^{2}}{R}$

$a_{c} = \frac{463.967^{2}}{6.38\times 10^{6}} = 0.0337\ m/s^{2}$

(b) To calculate the speed and acceleration at an altitude of $65.0^{\circ}$ N:

The horizontal component of the radius, R' = $Rcos\theta$

$R' = 6.38\times 10^{6}cos65.0^{\circ} = 2.69\times 10^{6}\ m$

Now,

For the speed, v' = $R'\omega = 2.69\times 10^{6}\times 7.27\times 10^{- 5} = 196.02\ m/s$

For the centripetal acceleration,

$a'_{c} = \frac{v'^{2}}{R'}$

$a'_{c} = \frac{196.02^{2}}{2.69\times 10^{6}} = 0.01428\ m/s^{2}$

8 0

4 years ago

A sample of an ideal gas is slowly compressed to one-half its original volume with no change in pressure. If the original root-m

Zigmanuir [339]

Answer:

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

Explanation:

From the kinetic theory model of gases we know that the velocity rms (speed of gas molecules) is given by:

$V_{rms}= \sqrt{\frac{3RT}{M}}$ (1)

Where V represent the velocity

R the constant for ideal gases

T the temperature

M the molecular weight of the gas

We also know from the ideal gas law that $PV= nRT$

If we solve for T we got: $T = \frac{PV}{nR}$

For the initial state we can replace T into the equation (1) and we got:

$V_{rms_i}= \sqrt{\frac{3R (\frac{PV}{nR})}{M}} = \sqrt{\frac{3PV}{M}}$

For the final state we know that : $V_f = \frac{V}{2}$ And the pressure not change , so then the final velocity would be:

$V_{rms_f}= \sqrt{\frac{3R (\frac{P(V/2)}{nR})}{M}} = \sqrt{\frac{3P(V/2)}{M}}$

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

6 0

3 years ago

What is the current I(3τ), that is, the current after three time constants have passed? The current in the circuit will approach

Olin [163]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$I(\tau)=0.051 A$

$I(3 \tau)=0.076 A$

$I_c= 0.08 A$

Explanation:

From the question we are told that

$I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R$

From the question we are told to find $I(\tau)$ when t=0 equals the time constant ( $\tau$ )

That is to obtain $I(\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{\tau}{\tau} })$

Substituting 12 V for $\epsilon$ and 150Ω for R

$I(\tau) = \frac{12}{150} (1- e^{-1})$

$=0.051 A$

From the question we are told to find $I(3 \tau)$ when t=0 equals the 3 times the time constant ( $\tau$ )

That is to obtain $I(3\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{3\tau}{\tau} })$

$I(\tau) = \frac{12}{150} (1- e^{-3})$

$=0.076 A$

As tends to infinity $\frac{\infty}{\tau} = \infty$

So $I_c$ would be mathematically evaluated as

$I_c=I(\infty) = \frac{12}{150} (1- e^{- \infty})$

$= \frac{12}{150}$

$= 0.08 A$

5 0

4 years ago

A spool is placed on table top. The static friction coefficient between the table top and the spool is high.

ZanzabumX [31]

The best option will be 4. In the opposite direction of the applied force. If we think about angular motion and take clockwise direction as negative and anti-clockwise as positive. By finding the direction that is clockwise Or negative you can say. So, the movement of the spool will be opposite in direction of the applied force.

4 0

4 years ago

Read 2 more answers