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frozen [14]
3 years ago
15

Question: You have an apple tree that produces high-quality apples in large numbers. The tree is also resistant to a number of h

armful conditions and pests. Which is the best way to grow more such trees in your orchard?
Options:

A. grafting
B. greenhouse technology
C. layering
D. tissue culture
Physics
2 answers:
serious [3.7K]3 years ago
8 0

Answer: B. greenhouse technology

Explanation:

Greenhouse technology is the technology which can be used to grow and cultivate the crops in closed glass house. This will trap the sunlight and allow to grow the crops in cold regions of the world. As the plants are remained protected from the external environment they will remain protected against the activity of the harmful pests.

Greenhouse will be the correct option to grow more apple trees as this technology will provide high quality apples, in large number due to increase in resistance with the pests.

Whitepunk [10]3 years ago
5 0
B. Greenhouse technology 

Since the green house will help keep the pest and harmful conditions. The greenhouse will be able to control temperatures and will keep out harmful bugs. You/I will be able to provide the tree with the perfect sequence of growth. 
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Can yall please help me with this?
Lostsunrise [7]

Answer: b

Explanation:

B is a chemical property

Reference what are physical properties

And what are chemical properties

Under certain conditions it can release a gas is the key words letting you know this is a chemical property

8 0
3 years ago
What kind of organisms are able to convert sunlight into usable energy?
Tamiku [17]

Answer:

Photosynthetic organisms

Explanation:

The electromagnetic energy of sunlight is converted to chemical energy in the chlorophyll-containing cells of photosynthetic organisms. In eukaryotic cells these reactions occur in the organelle known as the chloroplast

Hope this helps! :)

4 0
3 years ago
Read 2 more answers
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Please help me promise would mark you brainiest
Triss [41]
B. Please mark me the brilliantest pls
3 0
2 years ago
Formula: KE = 1/2 my?
9966 [12]

Answer:

22400 Joules

Explanation:

Apply the formula:

KE = 1/2 . 40 . 1120

KE = 20 . 1120

KE = 22400 Joules

3 0
2 years ago
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