Answer:
a) F_b = 6.62 N
b) F_net = 5.583 N
Explanation:
Given:
- Conditions of He gas: T = 0 C , P = 1 atm , ρ = 0.179 kg/m^3
- The mass of balloon m = 0.012 kg
- The radius of balloon r = 0.5 m
Find:
a)What is the magnitude of the buoyant force acting on the balloon?
b)What is the magnitude of the net force acting on the balloon?
Solution:
- The buoyant force F_b acting on the balloon is equal to the weight of the air it displaces.The mass of the displaced air ρ*V is the volume of the balloon times the density of the. Multiplying that by acceleration due to gravity gives its weight.
F_b = ρ*V*g
F_b = 4*ρ*g*pi*r^3 / 3
F_b = 4*1.29*9.81*pi*.5^3 / 3
F_b = 6.62 N
- The net force will be the difference between the balloon’s weight and the buoyant force. The weight of the balloon is the density of the helium times the volume of the balloon added to the mass of the empty balloon.
F_g = ρ*V*g + m*g
F_g = 4*ρ*g*pi*r^3 / 3 + 0.012*9.81
F_g = 4*0.179*9.81*pi*.5^3 / 3 + 0.012*9.81
F_g = 1.037 N
- The net force is the difference between weight and buoyant force
F_net = F_g - F_b
F_net = 6.62 - 1.037
F_net = 5.583 N
1. Volume
2. liquid
3. Gas
4. Kinetic energy?
5.Gas
Answer:
Explanation:
We shall apply law of conservation of mechanical energy for projectile being thrown .
Total energy on the surface = total energy at height h required
a ) At height h , velocity = .351 x ( 2 GM/R x h )



h = .14 R
b )


h = .54 R
c ) least initial mechanical energy required at launch if the projectile is to escape Earth
= GMm / R + 1/2 m (2GM/R)
= 0
Decrease the work being done and decrease the time in which the work is to be completed.
Answer:
a. The stored electric field energy can be greater than the stored magnetic field energy.
b. As the capacitor is discharging, the current is increasing.
d. The stored electric field energy can be less than the stored magnetic field energy.
e. The stored electric field energy can be equal to the stored magnetic field energy.
Explanation:
We know that total energy of capacitor and inductor system given as

L=Inductance ,I =current
C=Capacitance ,V=Voltage
The total energy of the system will be remain conserve.If energy in the inductor increases then the energy in the capacitor will decrease and vice -versa.
The values of stored energy in the capacitor and inductor can be equal ,greater or less than to each other.But the summation will be constant always.
Therefore the following option are correct
a ,b , d , e