Answer:
It would take
time for the capacitor to discharge from
to
.
It would take
time for the capacitor to discharge from
to
.
Note that
, and that
.
Explanation:
In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is
, and the initial charge of the capacitor be
. Then at time
, the charge stored in the capacitor would be:
.
<h3>a)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
<h3>b)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
Answer: 2561.7 pounds
Explanation:
If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:
(1)
Where:
is the slope of the line
is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)
is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)
This means we already have one point of the graph, which coordinate is:

Rewritting (1):
(2)
As Y is a function of X:
(3)
Substituting the known values:
(4)
(5)
(6)
Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):
(7)
(8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.
Answer:
Cruising at 35,000 feet in an airliner, straight toward the east,
at 500 miles per hour
Explanation:
Answer:
...do
Explanation:
24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5
cm. What is the length of the book in mm?
25. Explain the modifications
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as


= 0.42 m/s²