Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
Answer:
8.08 x 10^-5 m
Explanation:
A = 2 mm^2 = 2 x 10^-6 m^2
Total number of electrons, N = 9.4 x 10^18
time, t = 3 s
n = 5.8 x 10^28 electrons/ m^3
Current, i = Q / t = N x e / t = (9.4 x 10^18 x 1.6 x 10^-19) / 3 = 0.5 A
Let vd be the drift velocity.
i = n e A vd
0.5 = 5.8 x 10^28 x 1.6 x 10^-19 x 2 x 10^-6 x vd
vd = 2.7 x 10^-5 m/s
Distance traveled by the electrons = velocity x time
= vd x t = 2.7 x 10^-5 x 3 = 8.08 x 10^-5 m
Answer: 0.049 mol
Explanation:
1) Data:
n₁ = 0.250 mol
p₁ = 730 mmHg
p₂ = 1.15 atm
n₂ - n₁ = ?
2) Assumptions:
i) ideal gas equation: pV = nRT
ii) V and T constants.
3) Solution:
i) Since the temperature and the volume must be assumed constant, you can simplify the ideal gas equation into:
pV = nRT ⇒ p/n = RT/V ⇒ p/n = constant.
ii) Then p₁ / n₁ = p₂ / n₂
⇒ n₂ = p₂ n₁ / p₁
iii) n₂ = 1.15atm × 760 mmHg/atm × 0.250 mol / 730mmHg = 0.299 mol
iv) n₂ - n₁ = 0.299 mol - 0.250 mol = 0.049 mol
Answer:
0 m/s
Explanation:
velocity= change in displacement/ time
at rest, the ball does not travel any distance
0/ t
=0
Answer:
the answer is at the BOTTOM OF THEIR QUESTION
Explanation:
IT IS CORRECT BTW
