Answer:
a) h'= 5/7 h
, b) h ’= h
Explanation:
Let's use energy conservation for this exercise
Starting point. Upper left side
Em₀ = mg h
Final point. Lower left side
Emf = K = ½ m v² + ½ I w²
Em₀ = emf
mgh = ½ m v² + ½ I w²
Angular and linear velocity are related
v = r w
w = v / r
The moment of inertia of the marble that we take as a solid sphere is
I = 2/5 m r²
We substitute
m g h = ½ m v² + ½ 2/5 m r² (v / r)²
g h = ½ v2 (1 + 2/5)
v = √(g h 10/7)
This is the speed at the bottom of the bowl
Now let's apply energy conservation to the right side
a) right side if rubbing
Em₀ = K
Emf = U = mg h’
½ m v² = mg h’
h’= ½ (g h 10/7) / g
h'= 5/7 h
b) right side with rubbing
Em₀ = K
Emf = K + U = -½ I w² + m g h
Emf = -½ 2/5 m r² v² / r² + m gh
Em₀ = emf
½ v² = -1/5 v² + g h’
h’= (1/2 +1/5) (gh 10/7) / g
h ’= h
c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher
Answer:F/L = μ0*I²/2πd
Explanation:
Both wires carries current of the value (I) and are separated by the distance (d).
The length of both wires is (L).
The first wire creates a magnetic field of the magnitude according to Bio-savart law
B=(Uo * I)/2πd
This magnetic field has the ability to exert a force F on the second wire.
This force (F) is given as
F=BIL* sin θ
θ here is 90 because the magnetic field is perpendicular to the length of the second wire.
Thus the value of sin 90 = 1
Hence F= BIL
But B=(Uo * I)/2πd.
We have that F= B=[(Uo * I)/2πd] x IL
Thus we have that F= [Uo* I²/2πd] x L
By rearranging, and bringing L to the left hand side of the equation, we have that
F/L = [Uo* I²/2πd]
Answer:
Your mobile charger is an example of (d.) a generator.
Explanation:
Your mobile charger generates energy to charge your device.
Answer:
a) 
b) 
c) 
Explanation:
Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:
Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.
When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.
a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:
b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.
The new surface charge density can be calculated as follows:
c) The electric field outside the cylinder can be found by Gauss' Law:
We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.
3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy
