The solenoid consists of N turns. The change in magnetic flux of the solenoid is −7,9 Wb in a time interval of 4,76 s. The emf induced in the coil is 31,54 V.

To find the number of turns of wire \( N \) on the solenoid, we can use Faraday's law of electromagnetic induction, which states:

\[
\text{emf} = -N \frac{\Delta \Phi}{\Delta t}
\]

Where:
- \(\text{emf}\) is the induced electromotive force (31.54 V in this case),
- \(\Delta \Phi\) is the change in magnetic flux (-7.9 Wb),
- \(\Delta t\) is the time interval (4.76 s),
- \( N \) is the number of turns of the solenoid.

We can rearrange the formula to solve for \( N \):

\[
N = -\frac{\text{emf} \cdot \Delta t}{\Delta \Phi}
\]

Now we can plug in the values:

- \(\text{emf} = 31.54\) V
- \(\Delta t = 4.76\) s
- \(\Delta \Phi = -7.9\) Wb (note that we are interested in the absolute value of \(\Delta \Phi\) since it's squared in the calculation).

Now we need to substitute these values into the equation:

\[
N = -\frac{31.54 \cdot 4.76}{-7.9}
\]

This simplifies to:

\[
N = \frac{31.54 \cdot 4.76}{7.9}
\]

Now we calculate the values:

1. Calculate \( 31.54 \cdot 4.76 \):

\[
31.54 \cdot 4.76 \approx 150.0944
\]

2. Now divide by 7.9:

\[
N \approx \frac{150.0944}{7.9} \approx 18.97
\]

Since the number of turns \( N \) must be a whole number, we round to the nearest whole number:

\[
N = 19
\]

Thus, the number of turns of wire on the solenoid is approximately \( \boxed{19} \).