Answer:
Choose/Pick your topic which is required for your target audience. Do some research and collect the needed information for your selected topic. Organize all important topic related facts & stats in a logical way. Write all your reader’s needs to influence & help them. Make your views unique & specific. Read, Revise, and Repeat. Cross-check the grammatical mistakes & avoid unnecessary & repetitive lines
<em>Hope this helps!!</em>
Answer:
a) W = 6.75 J and b) v = 3.87 m / s
Explanation:
a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition
W = ∫ F. dx
Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product
W = ∫ F dx
We replace and integrate
W = ∫ (-60 x - 18 x²) dx
W = -60 x²/2 -18 x³/3
Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m
W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]
W = 7.5 - 0.75
W = 6.75 J
b) Work is equal to the variation of kinetic energy
W = ΔK
W = ΔK = ½ m v² -0
v =√ 2W/m
v = √(2 6.75/ 0.90)
v = 3.87 m / s
Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;
Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
lmoooooi9okkkkjghedjydthaksidhqelzyakx
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration 
directed downward, initial vertical position 
, and initial vertical velocity 
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
substituting, we find
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
In order to do that, we use again the same SUVAT equation substituting 
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
Which means that the heigth of the packet was
