1 inch = 2.54 centimeters
All we need to do is multiply.
68.5 * 2.54 = 173.99cm
Best of Luck!
Answer:
The gauge pressure is 
Explanation:
From the question we are told that
The height of the water contained is 
The height of liquid in the cylinder is 
At the bottom of the cylinder the gauge pressure is mathematically represented as

Where
is the pressure of water which is mathematically represented as

Now
is the density of water with a constant values of 
substituting values


While
is the pressure of oil which is mathematically represented as

Where
is the density of oil with a constant value

substituting values


Therefore


The statements of both students are incorrect.
-- Electrical power, just like mechanical power, is expressed in units of watts.
-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms' means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).
There are no 'following statements'.
All in all, a very disappointing question.
Answer:
(a) 0.0171 V
Explanation:
A = 0.09 m^2, dB/dt = 0.190 T/s
(a) According to the law of electromagntic induction
e = dФ / dt
e = A dB / dt
e = 0.09 x 0.190 = 0.0171 V
(b)
as we know
i = e / R
we can find induced current by dividing induced emf by resistance
B is your answer oscillating