Answer:
confocal microscopy
Explanation:
According to my research on different types of microscopes, I can say that based on the information provided within the question the tool being mentioned in this situation is a confocal microscopy. This is an extremely powerful microscope used to develop extremely sharp images of cells and tissues by viewing one plane of the specimen at a given time.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
The jnd for a 100-gram weight, according to Weber's law will be 10 gram.
<h3>What is Weber's law?</h3>
It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.
In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;
= 100 / 10
= 10 gram
Therefore, jnd for a 100-gram weight, according to Weber's law will be 10 grams.
Learn more about weight on:
brainly.com/question/19753744
#SPJ1
Answer:
<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>
Explanation:
Answer:
The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².
How many times stronger than gravity is this force? 2.79 x 10⁴ g.
Explanation:
given information:
s = 220 m
final speed, vf = 10.97 km/s = 10970 m/s
g = 9.8 m/s²
he unrealistically large acceleration experienced by the space travelers during their launch
vf² = v₀²+2as, v₀ = 0
vf² = 2as
a =vf²/2s
= (10970)²/(2x220)
= 2.7 x 10⁵ m/s²
Compare your answer with the free-fall acceleration
a/g = 2.7 x 10⁵/9.8
a/g = 2.79 x 10⁴
a = 2.79 x 10⁴ g
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.