Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Answer:
a) T² = (
) r³
b) veloicity the dependency is the inverse of the root of the distance
kinetic energy depends on the inverse of the distance
potential energy dependency is the inverse of distance
angular momentum depends directly on the root of the distance
Explanation:
1) for this exercise we will use Newton's second law
F = ma
in this case the acceleration is centripetal
a = v² / r
the linear and angular variable are related
v = w r
we substitute
a = w² r
force is the universal force of attraction
F = 
we substitute
w² = 
angular velocity is related to frequency and period
w = 2π f = 2π / T
we substitute
the final equation is
T² = () r³
b) the speed of the orbit can be found
v = w r
v = 
v = 
in this case the dependency is the inverse of the root of the distance
Kinetic energy
K = ½ M v²
K = ½ M GM / r
K = ½ GM² 1 / r
the kinetic energy depends on the inverse of the distance
Potential energy
U =
U = -G mM / r
dependency is the inverse of distance
Angular momentum
L = r x p
for a circular orbit
L = r p = r Mv
L =
L =
The angular momentum depends directly on the root of the distance
Explanation:
Given that,
Capacitor = 30μC
Resistor = 49.0Ω
Voltage = 30.0 V
Frequency = 60.0 Hz
We need to calculate the impedance
Using formula of impedance
.....(I)
We need to calculate the value of 
Using formula of
Put the value of into the formula of impedance
(a). We need to calculate the rms current in the circuit
Using formula of rms current
The rms current in the circuit is 0.30 A.
(b). We need to calculate the rms voltage drop across the resistor
Using formula of rms voltage
Put the value into the formula
The rms voltage drop across the resistor is 14.7 V
(c). We need to calculate the rms voltage drop across the capacitor
Using formula of rms voltage
The rms voltage drop across the capacitor is 26.53 V.
Hence, This is the required solution.
Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
= R* 
in(
/
)
The specific heat capacity:
=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
=1 - (
/
)
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67
Answer:
Uniform rectilinear movement (m. r. u.)
Explanation:
It is a continuous movement without acceleration, that is, it moves at a constant speed. The speed does not change over time, for this reason, there is no change in acceleration.
Vf = final velocity = 50/30 [km/s] = 1.67 [km/s]
Vo = initial velocity = 50/30 [km/s] = 1.67 [km/s]
