Answer:
Explanation:
The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:
where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes
T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:
w = mg and
5500 =- m(10) so
m = 550 kg. Now we have what we need to solve for the tension:
T = 550(4.0) + 5500 and
T = 2200 + 5500 so
T = 7700 N
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.
The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:

The potential is given by the magnitude of the electric field, E, times the distance, d:

So we have

(1)
However, the electric field is equal to the electrical force F divided by the charge q:

Therefore (1) becomes

And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
Answer:
cyguu and I will be like that and I
Explanation:
GHG and I will be like them and I will be like this in the best of luck in the pulley system and the same your time and consideration and I will be like them and I will be like them and I will be like them and TFT capacitive touchscreen and I will be in Phone
Answer:
C. to stretch joints and muscles regularly
Explanation: