Yes, because power is a measure of how quickly work is done. In this case, the power for the 200 joules of work done in 20 seconds would be 10 watts, while the power for the 50 joules of work done would be 2.5 watts. Hence, 200 joules of work done involves more power than 50 joules of work done.
The power required to do 60 joules of work in 20 seconds is determined by the formula: Power = Work / Time. Therefore, Power = 60 joules / 20 seconds = 3 watts. So, 3 watts of power is necessary to do 60 joules of work in 20 seconds.
Two different units that represent work are joules (J) and ergs (erg). Joules are a standard metric unit for work and energy, while ergs are a unit commonly used in physics and are equal to 10^-7 joules.
The heat generated by 1700 joules of work depends on the efficiency of the process. In an ideal case where all the work is converted into heat, the heat generated would also be 1700 joules. However, in real-world scenarios, the heat generated would be less due to energy losses.
The total work done by the student is the sum of the work done in each direction. The work done pushing the cart 3.0 meters east is 300 Joules, and the work done pushing the cart 4.0 meters north is 400 Joules. Therefore, the total work done by the student is 300 Joules + 400 Joules = 700 Joules.
Work is measured in Joules.
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Force times work doesn't give joules. Joules is a unit of energy or work. Perhaps you mean the relation: force x distance = work.
Work is energy, measured in joules. The rate of work, or joules per second, is known as watts, or power.
Yes, because power is a measure of how quickly work is done. In this case, the power for the 200 joules of work done in 20 seconds would be 10 watts, while the power for the 50 joules of work done would be 2.5 watts. Hence, 200 joules of work done involves more power than 50 joules of work done.
Work is measured in Joules.
Joules
10 joules of work in 1 second
Manufacturing. A machine has a work output of 14ax4 j (joules) and a work in put 7a2x5 j. What is the eficiency of the machine? efficiency in % = work output in joules work input in joules
The efficiency of the lever can be calculated as the ratio of the output work to the input work, multiplied by 100% to express it as a percentage. In this case, the efficiency would be 870 joules (output work) divided by 930 joules (input work), multiplied by 100% which equals 93.55%.
Work is energy, measured in joules. The rate of work, or joules per second, is known as watts, or power.
-70 Joules