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svp [43]
3 years ago
13

Seasonal changes bring about scenes like this one. Plant cells respond to changes in _________________ and as a result, photosyn

thesis slows and leaves change color.
Physics
2 answers:
NikAS [45]3 years ago
6 0

The answer is; light intensity & temp change

As winter approaches, the chlorophyll molecules reduce (by being broken down) and this is why leaves begin to turn orange as there are more red pigments (Xanthophyll) than the green pigment. This is because as winter approached, the light intensity decreases and it gets colder. Therefore, the plants begin to prepare for the harsh incoming weather. Most deciduous trees even begins to shed their leaves.


jekas [21]3 years ago
4 0
Light intensity & temp change 
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7 0
3 years ago
A convex lens can produce a real image but not a viral image<br> a. true<br> b. false
serious [3.7K]
The answer is true a convex lens can produce a real image but not a viral image
5 0
4 years ago
Match the correct sentence together.
Brut [27]
I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B
6 0
2 years ago
Read 2 more answers
A car is moving with a speed of 15 ms. How long does it take to cover a distance<br>of 1.2 km?​
fgiga [73]

Answer: 80 s

Explanation:

Speed is expressed in  v= d/t, derive the equation so we can have time.

First conert km into meters to cancel out both units and only seconds will remain.

1.2 km x 1000m/ 1km = 1200 m

t = 1200 m /15 m/s = 80 s

8 0
3 years ago
2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an
Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

\delta = \frac{Pl}{AE}

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

\delta = \frac{Pl}{AE}

\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}

\delta = 0.1973mm

Therefore the elongaton of the rod in a 200mm gage length is 0.1973mm

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

\upsilon = \frac{l_{a}}{l_{s}}

Where,

\upsilon =Poission's ratio

l_{a}= Lateral strain

l_{s}= Linear strain

\upsilon = \frac{\delta/d}{\delta/l}

0.3 = \frac{\delta/22}{0.1973/200}

0.3(\frac{0.1973}{200}) = \frac{\delta}{22}

\delta = 6.5109*10^{-3}mm

Therefore the change in diameter of the rod is 6.5109*10^{-3}mm

5 0
3 years ago
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