To solve this problem we will apply the concepts related to translational torque, angular torque and the kinematic equations of angular movement with which we will find the angular displacement of the system.
Translational torque can be defined as,
Here,
F = Force
d = Distance which the force is applied
At the same time the angular torque is defined as the product between the moment of inertia and the angular acceleration, so using the previous value of the found torque, and with the moment of inertia given by the statement, we would have that the angular acceleration is
Now the angular displacement is
Here
= Initial angular velocity
t = time
Angular acceleration
= Angular displacement
Time is given as 1 minute, in seconds will be
There is not initial angular velocity, then
Replacing,
The question neglects the effect of gravitational force.
Answer:
389.6 W/m²
Explanation:
The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K
So, P = σεA(T₁⁴ - T₂⁴)
h = P/A = σε(T₁⁴ - T₂⁴)
Substituting the values of the variables into the equation, we have
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴
h = 38955213360 × 10⁻⁸ W/m²
h = 389.55213360 W/m²
h ≅ 389.6 W/m²
Answer:
hL = 0.9627 m
Explanation:
Given
Q = 0.040 m³/s (constant value)
D₁ = 15 cm = 0.15 m ⇒ R₁ = D₁/2 = 0.15 m/2 = 0.075 m
D₂ = 8 cm = 0.08 m ⇒ R₂ = D₂/2 = 0.08 m/2 = 0.04 m
P₁ = 480 kPa = 480*10³Pa
P₂ = 440 kPa = 440*10³Pa
α = 1.05
ρ = 1000 Kg/m³
g = 9.81 m/s²
h₁ = h₂
hL = ? (the irreversible head loss in the reducer)
Using the formula Q = v*A ⇒ v = Q/A
we can find the velocities v₁ and v₂ as follows
v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s
v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s
Then we apply the Bernoulli law (for an incompressible flow)
(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL
Since h₁ = h₂ we obtain
(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL
⇒ hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)
⇒ hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)
⇒ hL = 0.9627 m
Answer:
The correct option is;
Loose cords
Explanation:
Based on the 2011 Census of Occupational Injuries, which is published by the Bureau of Labor Statistics, one of the leading causes of injuries at work is slips, trips and falls which may lead long duration of time down and large huge amount of claims for compensation. Slips trips and falls are also comes fourth in the reasons of fatality at work
Trip hazards are hazards that causes trip and fall by stopping and locking the movement of the step of people walking along traffic lanes
Items that cause trips mainly include item used for work. The correct option is therefore loose cords, which should be kept under cable bridges for safety.