The magnitude of emf induced in the single coil of wire rotated in the uniform magnetic field is 0.171 V.
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Induced emf</h3>
The emf induced in the loop is determined by applying Faraday's law of electromagnetic induction.
emf = N(dФ/dt)
where;
- N is number of turns of the wire
- Ф is magnetic flux
Ф = BA
where;
- B is magnetic field strength
- A is the area of the loop
emf = NBA/t
A = πr²
A = π x (0.13)²
A = 0.053 m²
emf = NBA/t
emf = (1 x 3.746 x 0.053)/(1.16)
emf = 0.171 V
Thus, the magnitude of emf induced in the single coil of wire is 0.171 V.
Learn more about induced emf here: brainly.com/question/13744192
Answer:
final temperature is 364.32 K
mass of air is 0.5423 kg
Explanation:
given data
pressure p1 = 380 kPa
volume v1 = 0.120 m³
temperature = 20°C = 20 + 273 = 293 K
pressure p2 = 3450 kPa
volume v2 = 5% increase
to find out
final temperature and the mass
solution
we consider here ideal gas
so equation is
p1v1 / t1 = p2v2/t2
put here all value and find t2
and v2 is = 0.120 ( 1 + 0.05) = 0.126 m³
so
t2 = p2v2/ p1v1 × t1
t2 = (450 × 0.126) / (380 × 0.120) × 293
t2 = 364.32 K
so final temperature is 364.32 K
and
mass = p2v2 / Rt2
here R gas constant is 0.287 kJ/kg.K
so
mass = (450 × 0.126) / (0.287 × 364.32 )
mass = 0.5423
so mass of air is 0.5423 kg
Answer:
Explanation:
First of all, let's convert from nanometres to metres, keeping in mind that
So we have:
Now we can convert from metres to centimetres, keeping in mind that
So, we find:
Answer: It is the ratio of solute in a solution to either solvent or total solution.
Explanation:
