They have a diagonal relationship
Answer:
B. 12.0A
Explanation:
The circuit shown in the image is a parallel circuit which mean that the voltage of the source is equal across each resistance in the circuit because they are directly connected to the terminals of the voltage source.
The voltage across the 10 Ohms resistor is 120V
So, let's proceed to calculate the current
By using Ohm's Law
12A is the amount of current flowing through the 10 ohms resistor.
If you press on your arm force is applied work done is if it moves.
One answer could be if I was to press my hand on a table.
Have a great day!
Answer:
The tension in string is found to be 188.06 N
Explanation:
For the vibrating string the fundamental frequency is given as:
f1 = v/2L
where,
f1 = fundamental frequency = 335 Hz
v = speed of wave
L = length of string = 28.5 cm = 0.285 m
Therefore,
v = f1 2L
v = (335 Hz)(2)(0.285)
v = 190.95 m/s
Now, for the tension:
v = √T/μ
v² = T/μ
T = v² μ
where,
T = Tension
v = speed = 190.95 m/s
μ = linear mass density of string = mass/L = 0.00147 kg/0.285 m = 5.15 x 10^-3 kg/m
Therefore,
T = (190.95 m/s)²(5.15 x 10^-3 kg/m)
<u>T = 188.06 N</u>
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.
