<span>Balloons are blown up, and then rubbed against your shirt many times. The balloon then touches the ceiling. When released, the balloon remains stuck to the ceiling. The balloon is charged by contact. The ceiling has a neutral charge. The charged balloon induces a slight surface charge on the ceiling opposite to the charge on the balloon. Balloon and ceiling electric charges are opposite in sign, so they will attract each other. Since both the balloon and the ceiling are insulators, charge can not flow from one to the other. The charge on the balloon is fixed on the balloon and the charge on the ceiling remains fixed to the ceiling. It just so happens that the<span> electrostatic force the ceiling exerts on the balloon is sufficient to hold the balloon in place (i.e. overcomes gravity, etc.).</span></span>
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:
3.9 m/s
Explanation:
We are given that
Mass of car,m=
Initial velocity,u=0
Distance,s=5.9 m
Average friction force,f=
We have to find the speed of the car at the bottom of the driveway.
Net force,
Where 
Acceleration,
v=3.9 m/s
