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3 years ago

Describe the type of energy transfers that occur in photosynthesis

Physics

1 answer:

Alexeev081 [22]3 years ago

4 0

Answer:

light energy transfers electrons from water (H 2 O) to carbon dioxide (CO 2), to produce carbohydrates. In this transfer, the CO 2 is "reduced," or receives electrons, and the water becomes "oxidized," or loses electrons

Explanation:

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There are three naturally occurring isotopes

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$^{44}\text{Hi}: 44.000\:\text{g/mol}×0.288 = 12.672\:\text{g/mol}$

Adding all the weighted values on the right hand side, we get

$\text{Hi}: 42.293\:\text{g/mol}$

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3 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

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3 years ago