Question

A vacuum cleaner has an energy input of 6.5 kWh and an energy output of 5.8 kWh. The efficiency of the vacuum cleaner, to the nearest percentage, is

Answer 1

Answer:

89%

Explanation:

Given:

Input energy to a vacuum cleaner $(E_{in})$ = 6.5 kWh

Output energy to a vacuum cleaner $(E_{out})$ = 5.8 kWh

The efficiency of a machine is a measure of useful work provided by the machine.

When a machine is operated, some of its input energy is lost in friction and other wear and tear. So, the total input energy is not converted to output energy. In order to measure this loss, we make use of the term "efficiency" which is defined as the ratio of output energy to input energy expressed as a percentage.

Thus, efficiency of a vacuum cleaner is given as:

$\eta=\frac{Out put\ energy}{In put\ energy}\times 100\\\\\eta=\frac{E_{out}}{E_{in}}\times 100$

Plug in the given values and solve for $\eta$. This gives,

$\eta=\frac{5.8\ kWh}{6.5\ kWh}\times 100\\\\\eta=0.8923\times 100\\\\\eta=89.23\approx 89\%$

So, 89% of the input energy is converted to useful work by the vacuum cleaner.

Therefore, the efficiency of the vacuum cleaner is 89%.

Answer 2

Answer:

The answer would be 89%

Explanation:

the equation you use is $\frac{energy out}{enrgy in} x 100=$ energy efficiency. You would then plug in the numbers with 5.8 on the to and 6.5 on the bottom, once you find the quotient multiply that by 100 which gives you 89.2307692308 and you have a percentage of 89%