The current flowing in each resistor of the circuit is 4 A.
<h3>
Equivalent resistance of the series resistors</h3>
The equivalent resistance of the series circuit is calculated as follows;
6 Ω and 4 Ω are in series = 10 Ω
5 Ω and 10Ω are in series = 15 Ω
<h3>Effective resistance of the circuit</h3>
<h3>Current flowing in the circuit</h3>
V = IR
I = V/R
I = 24/6
I = 4 A
Learn more about resistors in parallel here: brainly.com/question/15121871
Answer:
13.2m
Explanation:
Step one:
given data
Energy= 5610J
Force F= 425N
Required
The distance traveled
Step two:
We know that work done is given as
WD= force* distance
so
5610=425*d
divide both sides by 425
d= 5610/425
d=13.2m
Answer:
All points to the left of zero are negative
Explanation:
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
