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Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other.
, the amount of charge stored in this capacitor, will stay the same.
The formula
relates the electric potential across a capacitor to:
, the charge stored in the capacitor, and
, the capacitance of this capacitor.
While
stays the same, moving the two plates apart could affect the potential
by changing the capacitance
of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of
. Neither will that change the area of the two plates.
However, as
(the distance between the two plates) increases, the value of
will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula
can be rewritten as:
.
The value of
(charge stored in this capacitor) stays the same. As the value of
becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.
Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.
Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.
The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.
Learn more about grounding conductors here brainly.com/question/14886979
#SPJ4.
Answer:
16∠45° Ω
Explanation:
Applying,
Z = V/I................... Equation 1
Where Z = Impedance, V = Voltage output, I = current input.
Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°
Substitute these values into equation 1
Z = 120cos(10t+75°)/7.5cos(10t+30)
Z = 120∠75°/ 7.5∠30°
Z = 16∠(75°-30)
Z = 16∠45° Ω
Hence the impedance of the linear network is 16∠45° Ω
To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

g = Gravity
L = Length
T = Period
Re-arrange to find the gravity we have

Our values are given as

Replacing we have



Therefore the correct answer is C.