Answer: i would do something that is really easy and i will do test 2 where there aint a word problems and i am going 2 put the work 2 where it is 2 there standard i aint going 2 do sum over there standards of science.
Explanation:
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
Answer:
Speed greater than 4 m/s
Explanation:
Given that Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs to run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?
The distance = 12 m
Time = 3s
Speed = distance/time
Speed = 12/3
Speed = 4 m/s
Ms Kasper must run at speed more than 4m/s for her to save the cat.
