Answer:
Is the equation for Ec=1/2 m(Dv)^2 where Dv is the difference between the angular speed & the areolar speed?
The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.
Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.
The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.
To know more about working distance
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Answer:
The change in kinetic energy (KE) of the car is more in the second case.
Explanation:
Let the mass of the car = m
initial velocity of the first case, u = 22 km/h = 6.11 m/s
final velocity of the first case, v = 32 km/h = 8.89 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(8.89² - 6.11²)
= 20.85m J
initial velocity of the second case, u = 32 km/h = 8.89 m/s
final velocity of the second case, v = 42 km/h = 11.67 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(11.67² - 8.89²)
= 28.58m J
The change in kinetic energy (KE) of the car is more in the second case.
Answer:
8 electrons in the third energy level
Explanation:
From the description,the third energy level has 8 electron (represented by the small green balls you describe)
Answer: 1477.78 N
Explanation:
Let's assume that the cross sectional area of the smaller piston be A1
let's also assume the cross sectional area of the larger piston be A2
We assume the force applied to the smaller piston be F1
We also assume the force applied to the larger piston be F2
we then use the formula
F1/A1 = F2/A2
From our question,
The radius of the smaller piston is 5 cm = 0.05 m
The radius of the larger piston is 15 cm = 0.15 m
The force of the larger piston is 13300 N
The force of the smaller piston is unknown = F
A1 = πr² = 3.142 * 0.05² = 0.007855 m²
A2 = πr² = 3.142 * 0.15² = 0.070695 m²
F1/0.007855 = 13300/0.070695
F1 = (13300 * 0.007855) / 0.070695
F1 = 104.4715 / 0.070695
F1 = 1477.78 N
Thus, the force the compressed air must exert is 1477.78 N