Answer:
F = 39.2 N (hand force) and N = 68.6 N (shoulder force)
Explanation:
In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole
Let's set the reference system at the fit point of the shoulder
∑ τ = 0
We will assume that the counterclockwise turns are positive
w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0
all distances are measured from the support of the man (x₀ = 0.60 m)
F = (w₁ 0.60 + W 0.1) / 0.4
F = (m₁ 0.6 + m 0.1) g / 0.4
let's calculate
F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4
F = 39.2 N
this is the force that the hand must exert to keep the system in balance
We apply the translational equilibrium condition
-w₁ -W + N - F = 0
N = w₁ + W + F
N = (m₁ + m) g + F
let's calculate
N = (2.6 + 0.4) 9.8 + 39.2
N = 68.6 N
Answer:
ΔH°comb=-5899.5 kJ/mol
Explanation:
First, consider the energy balance:
Where 
is the calorimeter mass and 
is the number of moles of the samples; 
is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.
The product 
is the heat capacity, so the balance equation is:
So, the enthalpy of combustion can be calculated:
I will be happy to solve any doubt you have.
Answer:
Mass of another particle, m = 46 kg
Explanation:
it is given that,
Mass of first particle, m₁ = 28 kg
Gravitational force, 
Distance between the particles, d = 3.2 m
We need to find the mass m of another particle. It is given by the formula as follows :
m = 45.5 kg
or
m = 46 kg
So, the correct option is (d) "46 kg". Hence, this is the required solution.
Answer:
The banking angle of the road is 16.38 degrees.
Explanation:
Given:
Radius of the roadway on curve, R = 50 m
velocity of the moving car, V = 12 m/s
The banking angle can be calculated by using the formula below:
If there's no frictional force, then net force is equal to centripetal force.
MgTanΦ = (MV^2)/ R
TanΦ = V^2 / gR
Where;
Φ is the banking angle
g is acceleration due to gravity
TanΦ = (12 x 12) / (9.8 x 50)
TanΦ = 0.2939
Φ = arctan (0.2939)
Φ = 16.38 degrees
Therefore, the banking angle of the road is 16.38 degrees.
Answer:
<h2>
242.5kPa</h2>
Explanation:
According to one of the gas laws, 
Given P1 = 240.0kPa, T1, 15.0°C, T2 = 18.0°C, P2 = ?
Substituting this values into the equation, we have;
Cross multiplying we have;
288P2 = 240*291
The new pressure is 242.5kPa
