Answer:
Δ
= 84 Ω, = (40 ± 8) 10¹ Ω
Explanation:
The formula for parallel equivalent resistance is
1 / = ∑ 1 / Ri
In our case we use a resistance of each
R₁ = 500 ± 50 Ω
R₂ = 2000 ± 5%
This percentage equals
0.05 = ΔR₂ / R₂
ΔR₂ = 0.05 R₂
ΔR₂ = 0.05 2000 = 100 Ω
We write the resistance
R₂ = 2000 ± 100 Ω
We apply the initial formula
1 / = 1 / R₁ + 1 / R₂
1 / = 1/500 + 1/2000 = 0.0025
= 400 Ω
Let's look for the error (uncertainly) of Re
= R₁R₂ / (R₁ + R₂)
R’= R₁ + R₂
= R₁R₂ / R’
Let's look for the uncertainty of this equation
Δ / = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’
The uncertainty of a sum is
ΔR’= ΔR₁ + ΔR₂
We substitute the values
Δ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)
Δ / 400 = 0.1 + 0.05 + 0.06
Δ = 0.21 400
Δ = 84 Ω
Let's write the resistance value with the correct significant figures
= (40 ± 8) 10¹ Ω
I'm pretty sure the answer is gold
Answer:
Explanation:
We shall apply law of conservation of momentum to know velocity after collision . Let it be v .
total momentum before collision = total momentum after collision
15 x 1.5 - 12 x .75 = ( 15 + 12 ) v
v = .5 m /s
kinetic energy before collision
1/2 x 15 x 1.5² + 1/2 x 12 x .75²
= 16.875 + 3.375
= 20.25 J
kinetic energy after collision
= 1/2 x ( 15 + 12 ) x .5²
= 3.375 J
Loss of energy = 16.875 J
This energy appear as heat and sound energy that is produced during collision .
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
mλ
d = 1.0128×10⁻⁵m
Answer:
A
Explanation:
R=V/I
IF you double the resistance, it's become:
2R=(1/2)I
