Answer:
P = F/S = 100/2 =50 (N/m2)
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ
= m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s
Answer:
Explanation:
a = 4ms⁻², Vf = 180 m/s & Vi = 140m/s
a =
4 = 
t = 40/4
t = 10sec
To Measure Distance Use third Equation of Motion:
2aS = Vf²-Vi²
S = 
S = 12800/8 = 1600m
The lines can be traced out with a compass. The needle is like a permanent magnet and the north indicator is the north end of a magnet.
Answer:
the number of turns in the primary coil is 13000
Explanation:
Given the data in the question;
V₁ = 13000 V
V₂ = 120 V
N₁ = ?
N₂ = 120 turns
the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;
V₁/V₂ = N₁/N₂
V₁N₂ = V₂N₁
N₁ = V₁N₂ / V₂
so we substitute
N₁ = (13000 V × 120 turns) / 120 V
N₁ = 1560000 V-turns / 120 V
N₁ = 13000 turns
Therefore, the number of turns in the primary coil is 13000