Answer:
P = 1333.33 N
Explanation:
The pressure exerted by the boy on the floor can be calculated by the following equation:

where,
P = Pressure exerted by the boy = ?
F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)
A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)
A = 2(0.06 m x 0.25 m) = 0.03 m²
Therefore,

<u>P = 1333.33 N</u>
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.
Let V = the volume of the sample.
Then
W = ρVg
or
V = W/(ρg)
= (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
= 4.4972 x 10⁻⁵ m³
Answer: The volume is 44.972 x 10⁻⁶ m³
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i (
)
1 = i (
)
1 = i (
) =
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ =
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ =
R₂ =
R₂ = 19.9 Ω