Answer:
a) Q1=Q2=480μC V1=240V V2=60V
b) Q1=96μC Q2=384μC V1=V2=48V
c) Q1=Q2=0C V1=V2=0V
Explanation:
Let C1 = 2μC and C2=8μC
For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:




For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:
So, 
Total charge is the same calculated for part (a), so:
Solving for Q2:
Q2 = 384μC Q1 = 96μC.
Therefore:
V1=V2=48V
For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C V1=V2=0V
Velocity is the displacement of an object during a specific unit of time. Explanation: The standard unit of velocity is metre per second.
HOPE IT HELPS
Answer:
The initial temperature of the hot water is
(assuming that no heat was lost to the surroundings.)
Explanation:
Let
denote the mass of the hot water.
The question states that the mass of the water at
is three times the mass of the hot water. If the mass of the hot water is
, the mass of the cold water would be
.
Let
denote the specific heat capacity of water. Let
denote the mass of some water. The energy required to change the temperature of that much water by
(without state change) would be:
.
The temperature change for the cold water was:
.
Energy required to raise the temperature of water with mass
from
to
:
.
On the other hand, if the initial temperature of the hot water is
(where
,) the temperature change would be:
.
Calculate the energy change involved:
.
If no energy was lost to the surroundings,
should be equal to
. That is:
.
Simplify and solve for
:
.
.
Therefore, the initial temperature of the hot water would be
.
Explanation:
the third answer is correct
71.5m/s
Answer:
one truck load is 2.5*1.5*1 = 3.75m^3
7/3.75 = 1.86 loads,
so 2 trips must be made.
Explanation: