By law of refraction we know that image position and object positions are related to each other by following relation

here we know that



now by above formula


so apparent depth of the bottom is seen by the observer as h = 3.39 cm
I could be wrong, but I'm pretty sure it's 144kg.
Answer:
Gravity
Explanation:
The answer is gravity because when the 3 masses were hung from the spring, gravity pulled the spring towards the ground.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + (
-
)² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL