Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
<h3>
B. True</h3>
"This was the idea that non-living objects can give rise to living organisms."
As the roller coaster speeds up on the way down the hill, the potential energy of roller coaster will be converted to kinetic energy.
<h3>
What is Conservation of Energy ?</h3>
Conservation of energy state that energy is neither created nor destroy, they can only be transformed from one form to another. Energy of and object can transform from Potential energy to kinetic energy and vice versa
Given that at the top of a hill a roller coaster has gravitational potential energy due to its position. What will happen to this potential energy as the roller coaster speeds up on the way down the hill is that the potential energy to the roller coaster will start decreasing while the kinetic energy will start to increase.
The total energy of the roller coaster will be constant because of conservation of energy. As the roller coaster speeds up on the way down the hill, the potential energy will eventually reduce to zero where the total energy of the as the roller coaster will be equal to maximum kinetic energy.
Therefore, as the roller coaster speeds up on the way down the hill, the potential energy of roller coaster will be converted to kinetic energy.
Learn more about Energy here: brainly.com/question/25959744
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This question apparently comes after an EARLIER one,
where you were told either the voltage across the same
capacitor or the total charge stored in it. You can't answer
THIS one without that information.
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
