Answer
The frequency and the period are 45 cycles/second and 0.023 seconds
Explanation:
The frequency of the engine is given by :
[tex]f=\dfrac{rpm}{60}[/tex]
We have, the engine increases its performance from zero to 2700 rpms.
Frequency,
[tex]f=\dfrac{2700}{60}\ \text{cycles per second}\\\\f=45\ \text{cycles per second}[/tex]
The time period of the engine is given by :
T = f/f
So,
[tex]T=\dfrac{1}{45}\ s\\\\T=0.023\ s[/tex]
Hence, the required frequency and the period are 45 cycles/second and 0.023 seconds respectively.
