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3 years ago

14

Some trees have red leaves all year long. You know that plants with green leaves make their food for themselves. How do you thin

k trees with red leaves gets its food?

2 answers:

nexus9112 [7]3 years ago

8 0

"The chlorophyll needed for photosynthesis is 'hiding' within the leaf color, whether it be purple, yellow or red<span>. Our eyes lack the ability to see that chlorophyll is there." Plant </span>leaves<span> have three primary classes of pigments: chlorophyll, carotinoids and anthocyanins</span>

marshall27 [118]3 years ago

6 0

The same way as the trees with green leaves. Through photosynthesis :)

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Magnetic induction is used for what?

DochEvi [55]

For radio broadcasting, in electricity meters, in any generator.

4 0

3 years ago

The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally

dalvyx [7]

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= \frac{1}{2}mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m \frac{v^2}{r}$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31} kg)(6.95 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(3.00 \cdot 10^{-5} T)}=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = \frac{v^2}{r}= \frac{(6.95 \cdot 10^7 m/s)^2}{13.18 m}=3.66 \cdot 10^{14} m/s^2$

5 0

3 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

4 0

3 years ago

Give an example of how electrical energy and thermal energy are connected.

mafiozo [28]

A toaster draws electric current (electrical energy) from a wall outlet and converts these moving electric charges into heat (thermal energy) in the filaments that turn red hot to cook your toast. ... As electricity runs through the filaments in a space heater, the electrical energy is converted into heat (thermal energy).
Hopefully this helped.

3 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago