Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
- W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
= K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
Answer: More work will be done in case A
Explanation:
Work is said to be done when a force applied to an object causes the object to move through a distance or height.
Work = Force × distance
If the heavy load is pulled straight up with a massless rope, then the object is pulled through a height at an angle perpendicular to the ground (i.e 90°)
Much work will be done in case A and less work in case B. Less work is done in case B due to the use of a machine which is an inclined plane. The use of machine makes work much easier and it allows us to do work in a more convenient manner. It allow us to overcome a much larger load using minimal effort. For the first case, much work will be done since we are carrying through a vertical height which requires more effort compare to when an inclined plane is used.
Also note that the higher the angle of inclination, the more the work done on the object and vice versa
Disk of dust and gas contains the new born stars and as well as young stars and the Halo region contains the old stars.
The different regions of a spiral galaxy like our Milky way tend to have characteristic populations of stars.
Halo: The halo tends to have older, red stars. In general, blue stars are larger but since they are also hotter, tend to run through their fuel faster and have shorter lives. Halo stars tend to be older stars, The halo is generally lacking in dust and low in metals.
Bulge: The outer Area of the bulge has mostly older red stars. The inner bulge has active star formation and has a mix of red and blue stars.
Disk: The disk has active star formation regions, so there are young blue stars in that region. Here are the stars that are both old stars and young stars. The blue stars are largely found on the edges of the spiral arms.
Thus disk with the dust clouds contains new and young stars. And halo region contains old stars.
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Answer:The acceleartion is halved
Explanation:
This happens as acceleration is inversely proportional to the mass of an object according to Newton's 2nd law. So if the mass is doubled then the acceleration is halved