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Answer:
the work done by the lawnmower is 236.14 J.
Explanation:
Given;
power exerted by the lawnmower engine, P = 19 hp
time in which the power was exerted, t = 1 minute = 60 s.
1 hp = 745.7 watts
The work done by the lawnmower is calculated as follows;

Therefore, the work done by the lawnmower is 236.14 J.
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.
We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.

where,
- c: specific heat capacity of water
- m: mass of water
- ΔT: change in the temperature
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
Learn more: brainly.com/question/16104165
Answer:
the final speed of the rain is 541 m/s.
Explanation:
Given;
acceleration due to gravity, g = 9.81 m/s²
height of fall of the rain, h = 9,000 m
time of the rain fall, t = 1.5 minutes = 90 s
Determine the initial velocity of the rain, as follows;

The final speed of the rain is calculated as;

Therefore, the final speed of the rain is 541 m/s.