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
Answer: the object transferred some of its energy to its surroundings.
Explanation:
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True
2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6
3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
A) Agreed.
<span>b) Value agreed but units should be W (watts). </span>
<span>c) Here's one method... </span>
<span>15 miles = 24140 m </span>
<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>
<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>
<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>
<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>
<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>
<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
The area is the length times the height so A= bh ...?
