We are given an object that is speeding up on a level ground.
Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.
The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.
Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.
We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.
Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
Answer:
W = 8.92 10² kJ
Explanation:
For this exercise they give us the strength, we must calculate the distance traveled, for this we need the rocket acceleration let's use Newton's second law
F = m a
a = F / m
a = 20 103/1400
a = 14.29 m/s²
With kinematics we can find the distance traveled
² = v₀² + 2 a x
x = ( ²-v₀²) / 2 a
x = (50² -35²) / 2 14.29
x = 1275 / 28.58
x = 44.61 m
Let's calculate the work
W = F.d
The bold is vector; as indicated by the force is in the direction of movement the scalar product is reduced to the ordinary product
W = F d
W = 20 10³ 44.61
W = 8.92 10⁵ J
W = 8.92 10² kJ
Answer:
Explanation:
Given data
Length L=2.5 m
Radius R=d/2=30/2 = 15 mm
Torque based on allowable stress
Allowable shear stress τ=50 Mpa
Allowable torque T=(π/2)τc³
Torque based on allowable angle of twist
Allowable Angle of twist
Ф=7.5°
Ф=7.5×(π/180)=130.90×10⁻³ rad
Allowable torque
T=(GJФ)/L
T=(G(π/2)c⁴)Ф)/L
T=(πGc⁴Ф)/2
Maximum Power Transmitted
Maximum power transmitted is given by
