Answer:
Explanation:
Given that,
Current measure is
i=10±0.6 Amps
And also,
R=45.0±2.0 Ω
Power dissipated by
P=i²R
Then
P=(10±0.6)²(45.0±2.0)
P=10²×45
P=450Watts
Now, calculating the uncertainty
∆P=|P| • √(2(∆i/i)²+(∆R/R)²)
∆P=450√ (2×(0.6/10)²+(2/45)²)
∆P=450√(0.0072+0.001975)
∆P=450√0.009175
∆P=43.1
The uncertainty in power is 43.1
Then,
P=450 ± 43.1 Watts
Answer:
a) N₁ = 14083 turns, b) I₁ = 2.58 A
Explanation:
The relationship that describes the relationship between the primary and secondary of the transformer is
a) They indicate that the secondary has N2 = 130 turns, the turns of the primary are
N₁ =
N₁ =
N₁ = 14083 turns
b) since there are no losses, the power of the neighboring transformer is
P = V I
P = 120 280
P = 33600 W
this is the same power of the substation
P = V₁ I₁
I₁ = P / V₁
I₁ = 33600/13000
I₁ = 2.58 A
Answer:88711000joules
Explanation:
Firstly we have to convert 161km to meters
161km=161×1000m=161000m
Work=force×distance
Work=551×161000
Work=88711000joules
<span>Answer:
Sodium gives off a yellow flame. The sodium yeloow flame can obscure the faint lilac flame of potassium. So what you do is perform the test again, holding a square of cobalt glass to your eye to observe the flame. The cobalt glass screens out the intense sodium yellow, and you see the lilac flame of potassium loud and clear.</span>