Answer:
Output voltage is 1.507 mV
Solution:
As per the question:
Nominal resistance, R = 
Fixed resistance, R =
Gauge Factor, G.F = 2.01
Supply Voltage, 
Strain, 
Now,
To calculate the output voltage, 
:
WE know that strain is given by:
Thus
Now, substituting the suitable values in the above eqn:
Answer:
Explanation:
a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)
work done raised the potential energy
b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W
c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%
Not a very likely scenario.
Answer:
You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can't seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can't hurt or kill you.
Explanation:
You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can't seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can't hurt or kill you.
A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance
Here is my work, I find making charts like this one to find knowns and unknowns can be helpful
Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
