Answer:
- increasing use of hybrid crops
- altering genes in DNA to create new plants
- developing disease or pest resistant crops
Explanation:
The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.
For example,
- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.
- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.
- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.
Answer:
Equilibrium temperature will be 
Explanation:
We have given weight of the lead m = 2.61 gram
Let the final temperature is T
Specific heat of the lead c = 0.128
Initial temperature of the lead = 11°C
So heat gain by the lead = 2.61×0.128×(T-11°C)
Mass of the water m = 7.67 gram
Specific heat = 4.184
Temperature of the water = 52.6°C
So heat lost by water = 7.67×4.184×(T-52.6)
We know that heat lost = heat gained
So 
Electric force depends on the charge and the strength of the electric field. The equation that relates the three:
F = Eq where q is the charge and E is the electric field strength.
Answer:
The size of the image is 1.04 m.
Explanation:
Given that,
Height of object = 2.40 m
Distance of object = 2.60 m
Radius of curvature =4.00 m
Focal length 
We need to calculate the image distance
Using mirror formula
We need to calculate the height of the image
Using formula of magnification
Put the value into the formula
Hence, The size of the image is 1.04 m
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0
