A machine’s efficiency is determined to be 60%. The input work is 3,000 joules. What is the output work?(1 point)
Responses
50 joules
50 joules
5,000 joules
5,000 joules
180,000 joules
180,000 joules
1,800 joules
The output work of the machine can be found by multiplying the efficiency by the input work.
Output work = Efficiency * Input work
Output work = 0.60 * 3,000
Output work = 1,800 joules
Therefore, the correct answer is 1,800 joules.
A gardener uses a wheelbarrow as a lever to lift some dirt. The energy applied by the gardener is 2,000 joules. The efficiency of the work done is 95%. How much work is done on the dirt?(1 point)
Responses
1,900 joules
1,900 joules
2,105 joules
2,105 joules
2,000 joules
2,000 joules
1,905 joules
The work done on the dirt can be found by multiplying the efficiency by the energy applied by the gardener.
Work done on the dirt = Efficiency * Energy applied by gardener
Work done on the dirt = 0.95 * 2,000
Work done on the dirt = 1,900 joules
Therefore, the correct answer is 1,900 joules.
A pulley system is operating at only 50% efficiency. How could the efficiency be increased? (1 point)
Responses
by reducing the friction on the pulleys
by reducing the friction on the pulleys
by adding more pulleys to the system
by adding more pulleys to the system
by decreasing the amount of weight being lifted
by decreasing the amount of weight being lifted
by having a second person pulling on the rope
The correct answer is: by reducing the friction on the pulleys.
By reducing the friction on the pulleys, less energy would be lost to friction, resulting in an increase in efficiency.
What is the name of the fixed point about which a lever rotates?(1 point)
Responses
pulley
pulley
bar
bar
fulcrum
fulcrum
groove
The correct answer is fulcrum. The fulcrum is the fixed point about which a lever rotates.
Which car has the greatest efficiency?(1 point)
Responses
a 900 kg car that travels 15 km per liter of fuel
a 900 kg car that travels 15 km per liter of fuel
a 1,200 kg car that travels 20 km per liter of fuel
a 1,200 kg car that travels 20 km per liter of fuel
a 900 kg car that travels 20 km per liter of fuel
a 900 kg car that travels 20 km per liter of fuel
a 1,200 kg car that travels 15 km per liter of fuel
To determine the car with the greatest efficiency, we need to calculate the fuel consumption per kilogram of the car.
For the first car:
Fuel consumption per kilogram = 15 km/liter / 900 kg = 0.01667 km/kg
For the second car:
Fuel consumption per kilogram = 20 km/liter / 1,200 kg = 0.01667 km/kg
For the third car:
Fuel consumption per kilogram = 20 km/liter / 900 kg = 0.02222 km/kg
For the fourth car:
Fuel consumption per kilogram = 15 km/liter / 1,200 kg = 0.0125 km/kg
Comparing the values, we can see that the fourth car, a 1,200 kg car that travels 15 km per liter of fuel, has the greatest efficiency.
Therefore, the correct answer is a 1,200 kg car that travels 15 km per liter of fuel.
To determine the output work of a machine, we can use the formula for efficiency:
Efficiency = (Output Work / Input Work) * 100%
Given that the efficiency of the machine is 60% and the input work is 3,000 joules, we can rearrange the formula to solve for the output work.
Efficiency = (Output Work / 3,000 joules) * 100%
60% = (Output Work / 3,000 joules) * 100%
To solve for the output work, we can first convert the percentage to a decimal by dividing it by 100.
0.60 = (Output Work / 3,000 joules)
Next, we can isolate the Output Work by multiplying both sides of the equation by 3,000 joules.
Output Work = 0.60 * 3,000 joules
Simplifying the equation gives us:
Output Work = 1,800 joules
Therefore, the correct answer is 1,800 joules.