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.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
<u>FOR BULB 1:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
<u>FOR BULB 2:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
<u>A₁/A₂ = 0.44</u>
Answer:
power requirement is 23.52 × 
W
Explanation:
given data
flow rate q = 2 m³/s
elevation h = 1200 m
density of the water ρ = 1000 kg/m³
to find out
power requirement
solution
we will get power by the power equation that is
power = ρ× Q× g× h ...................1
put here all value we get power
power = ρ× Q× g× h
power = 1000 × 2 × 9.8 × 1200
power = 23.52 ×
so power requirement is 23.52 × W
Explanation:
Take north to be positive and south to be negative.
a = (v − v₀) / t
a = (-4.5 m/s − 4.5 m/s) / 8 s
a = -1.125 m/s²
The acceleration is 1.125 m/s² south.
Answer:
33.5 kJ
Explanation:
here there is no difference is made in the temperature. Only thing happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.
we have ice 100g =0.1 kg
Appplying Q=mL
Q= 
Q = 33 500 J
Q = 33.5 kJ
