Recall this equation for a device in a direct current circuit:
P = IV
P is the power dissipated by the device, I is the current through the device, and V is the voltage drop of the device.
If we choose to use the ampere as the unit of current and the volt as the unit of voltage, then the product of the current and the voltage will give the power with watts as the unit.
Answer:
Half life of S = 3.76secs
Explanation:
The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.
For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564
for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615
Simply say = 1/3.9 + 1/104 = 0.266015
So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components


Acceleration components




Time taken is 0.26088 seconds

The distance the beetle travels on the ground is 0.3677181864 m
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m
The initial volume of the gas is

while its final volume is

so its variation of volume is

The pressure is constant, and it is

Therefore the work done by the gas is

where the negative sign means the work is done by the surrounding on the gas.
The heat energy given to the gas is

And the change in internal energy of the gas can be found by using the first law of thermodynamics:

where the positive sign means the internal energy of the gas has increased.