<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: 
Explanation:
Given
Mass of child 
speed of child is 
Moment of inertia of merry go round is 
radius 
Conserving the angular momentum
Answer:
C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car.
Explanation:
A) According to the fourth law of thermodynamics, the temperature of the other parts of the car increases due to the coolant used for the engine.
B) According to the first law of thermodynamics, the hood of the car heats up using heat from the surroundings in-order to achieve thermal equilibrium with the engine.
C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car.
D) According to the third law of thermodynamics, the increase in the velocity of the car changes the entropy of the tires. To balance this change, the temperature of the other parts is increased.
Answer:
Explanation:
25 mm diameter
r₁ = 12.5 x 10⁻³ m radius.
cross sectional area = a₁
Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa
a )
velocity of blood v₁ = .6 m /s
Cross sectional area at blockade = 3/4 a₁
Velocity at blockade area = v₂
As liquid is in-compressible
a₁v₁ = a₂v₂
a₁ x .6 m /s = 3/4 a₁ v₂
v₂ = .8m/s
b )
Applying Bernauli's theorem formula
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²
13328 +190.8 = P₂ + 339.2
P₂ = 13179.6 Pa
= 13179 / 13.6 x 10³ x 9.8 m of Hg
P₂ = .09888 m of Hg
98.88 mm of Hg
Answer:
d) It will be cut to a fourth of the original force.
Explanation:
The magnitude of the electrostatic force between the charged objects is
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, the initial distance is doubled, so
r' = 2r
Therefore, the new electrostatic force will be
So, the force will be cut to 1/4 of the original value.
