Answer:
60 m/s
Explanation:
From the law of conservation of energy,
The kinetic energy of the plane = Energy of store in the spring when the plane lands.
1/2mv² = 1/2ke²
making v the subject of the equation.
v = √(ke²/m).................... Equation 1
Where v = the plane landing speed, k = spring constant, e = extension. m = mass of the plane.
Given: m = 15000 kg, k = 60000 N/m, e = 30 m.
Substitute into equation 1
v = √(60000×30²/15000)
v = √(4×900)
v = √(3600)
v = 60 m/s.
Hence the plane's landing speed = 60 m/s
Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
Answer:
The colors of the sunset result from a phenomenon called scattering. Molecules and small particles in the atmosphere change the direction of light rays, causing them to scatter. ... The short-wavelength blue and violet are scattered by molecules in the air much more than other colors of the spectrum.
Explanation:
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:


We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:

And the velocity achieved in part 1:

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

Then, to get
, we do:



And we substitute the values:

There are three types: divergent, convergent, and transform boundaries. I hope this helps.