When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load need
1 answer:
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The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.
Δm = ∑
Physical quantities are precise values of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.
When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the cases, all the errors add up.
If m is the calculated quantity, x_i the measured values and Δx_i the uncertainty of each value, the total uncertainty is
Δm = ∑ | dm / dx_i | Dx_i
for instance:
If the magnitude is a average of two magnitudes measured each with a different error
m =
Δm = | | Δx₁ + |
| Δx₂
= ½
= ½
Δm = Δx₁ + ½ Δx₂
Δm = Δx₁ + Δx₂
In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.
Learn more about propagation errors here:
Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F = C
ρV²A = W
we substitute
0.45 × × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F = T = C
ρV²A
we substitute
= 0.065 × × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW
Answer:
kE=0.0735 J
Explanation:
Given that
Radius ,R=10 cm = 0.1 m
Mass ,m= 3 kg
Angular speed ,ω = 3.5 rad/s
We know that moment of inertia for solid sphere given as
Kinetic energy
Now by putting the values
kE=0.0735 J
Therefore the kinetic energy will be 0.0735 J