Answer:
Explanation:
Plotting the original location of the helicopter before it flies 25 km north, it would be at the origin, (0,0) then after it flies north, the y vertex gains 25 points, so it would be (0,25)
After it flies east, the x coordinate gains 5 points, so it would now be (5,25)
After it flies south, the y coordinate loses or is subtracted by 5 points. so it would now be (5,20)
After flying west, the x coordinate loses 15 points. So the final vertex would be at (-10,20)
East = Right
West = Left
South= Down
North = Up
I used mainly mathematical methods by adding and subtracting the x and y coordinate values, but this could be graphed easily since I gave the coordinates just incase!
Hope this helps!
Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be 
and its mean distance of from the sun be 
.
Also let the orbital period of the planet be 
and its mean distance from the sun be 
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 × 
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 × )²×ω²× 34.64 ) / 2
0.050 ×(6.43 × )²×ω²× 34.64 = 108
ω² = 108 / 7.160 × 
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
The particles can undergo small oscillations around x₂.
The given parameters;
- <em>initial energy of the particles = E₁</em>
- <em>final energy of the particles, E₂ = 0.33E₁</em>
The movement of the particles depends on the kinetic energy of the particles.
When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.
However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.
- The maximum position the particle can oscillate is x₅
- The half position the particles can oscillate is x₃
Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.
Thus, we can conclude that the particles can undergo small oscillations around x₂.
Learn more here:brainly.com/question/23910777
