Answer:
The 10X objective is use for the identification of actual size of histology tissues and 4X magnification is best for observation of most tissues slides
Explanation:
4X magnification is best for observation of most tissues slides because it has an objective lens that have lower power and have great high field overview which make it very easier to locate specimens on the slide. It is use to get the overview of histology slides. It is use to showcase more detailed observations about histology.
The 40X objective is use majorly to identify tissue , to observe the finer details and study tissue organization on the histology slide.
Hi,
The answer is D, Gravitational.
Hope this helps.
r3t40
Answer :First part is what you learned so write what you learned then the other parts is The evidence that the universe is expanding comes with something called the red shift of light. Light travels to Earth from other galaxies. As the light from that galaxy gets closer to Earth, the distance between Earth and the galaxy increases, which causes the wavelength of that light to get longer.
Explanation:
Answer:
Therefore,
The magnitude of the force per unit length that one wire exerts on the other is

Explanation:
Given:
Two long, parallel wires separated by a distance,
d = 3.50 cm = 0.035 meter
Currents,

To Find:
Magnitude of the force per unit length that one wire exerts on the other,

Solution:
Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

where,

Substituting the values we get


Therefore,
The magnitude of the force per unit length that one wire exerts on the other is
