Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V
Answer:
v = - 1,715 m / s
, x = 0.0156 m
Explanation:
This is an oscillatory movement exercise, which is described by the expression
x = A cos (wt + Ф)
we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero
As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration
v = dx / dt
v = - A w sin wt
a = - Aw² cos wt
a = -g
-g = - Aw² cos wt
wt = cos⁻¹ (g / Aw²)
t = 1 / w cos⁻¹ (g / Aw²)
angular velocity and frequency are related
w = 2π f
w = 2π 4
w = 8π rad / s
remember that the angles are in radians
t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))
t = 0.039789 1.3473
t = 0.0536 s
let's find the speed for this time
v = - A w sin wt
v = - 0.07 8π sin (8π 0.0536)
v = - 1,715 m / s
the distance is
x = A cos wt
x = 0.07 cos (8π 0.0536)
x = 0.0156 m
Answer: most effective way is to practice reduce reuse and recycle for utilisation of resources
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
