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natulia [17]
3 years ago
14

What does a plant use in photosynthesis and what are the products? PLEASE HELP:)

Physics
1 answer:
LenaWriter [7]3 years ago
7 0

During photosynthesis: Light energy is absorbed by chlorophyll, a green substance found in chloroplasts in some plant cells and algae. Absorbed light energy is used to convert carbon dioxide (from the air) and water (from the soil) into a sugar called glucose. Oxygen is released as a by-product.


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An automobile engine has an efficiency of 19.0%. If it produces 23.0 kJ of mechanical work per second, the heat rejected per sec
creativ13 [48]

Answer:

(e) 98,1 KJ

Explanation:

The engine produces 19%; it means, it rejects 81% of energy. ⇒ 81/19=4.26 times.

The engine produces 23 kJ; it means it rejects 23 * 4.26 = 98.05263 kJ

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3 years ago
What is the velocity of a quarter drooped from a towards after 10secounds
ArbitrLikvidat [17]
U=0 
<span>t=10 </span>
<span>a=9.8m/s/s </span>
<span>v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) </span>

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3 years ago
A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

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3 years ago
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Ne4ueva [31]

Answer:

Chemical composition, Temperature, Radial velocity, Size or diameter of the star, Rotation.

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Elemental abundances are determined by analyzing the relative strengths of the absorption lines in the spectrum of a star.

The Spectral class to which the star belongs gives the information related to the temperature of the star. It is the spectral lines that determine the spectral class O B A F G K M are the spectral classes.

By measuring the wavelengths of the lines in the star's spectrum gives the radial velocity. Doppler shift is the method used to find the radial velocity.

A star can be classified as a giant or a dwarf . A giant star will have narrow width spectral lines whereas a dwarf star has wider spectral lines.

Broadening of the spectral lines will determine the star's rotation.

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