An airplane undergoes the following displacements: First, it flies 40 km in a direction 30° east of north. Next, it flies 56 km
due south. Finally, it flies 100 km 30° north of west. Using analytical methods, determine how far the airplane ends up from its starting point.
1 answer:
Answer:
Distance from start point is 72.5km
Explanation:
The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:
Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:
Replacing the given values in the equations, the distance is calculated.
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Refer to the diagram shown.
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
Answer:
The angular velocity is
Explanation:
Generally the acceleration experienced by the propeller blade's is broken down into
The Radial acceleration which is mathematically represented as
And the Tangential acceleration which is mathematically represented as
The net acceleration is evaluated as
Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture
So at maximum angular acceleration we a have
Explanation:
Given that,
Magnitude of vector A, |A| = 15
Magnitude of vector B, |B| = 25
We need to find the magnitude of this sum.
The maximum sum of the resultant vector,
The minimum sum of the resultant vector,
So, the magnitude of this sum either 45 or -10.