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When work is done by a machine, the fundamental principles of physics governing the conservation of energy and conservation of momentum remain unchanged. These laws dictate that energy cannot be created or destroyed, only transferred or converted, and that momentum will be conserved in any isolated system.
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How much is power output of a machine that lifts a 500kg object through a height of 20m in a time of 60s?
The work done by the machine would be the force required to lift the object multiplied by the distance it was lifted, which is 500kg * 9.81 m/s^2 * 20m = 98100 J. The power output would then be work done divided by time taken, which is 98100 J / 60 s = 1635 W or 1.64 kW.
A 50kg crate is lifted to a height of 10 meters if it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
The work done to lift the crate is equal to the gravitational potential energy gained: Work = force x distance = weight x height. Here, Work = 50kg x 9.8m/s^2 x 10m = 4900 Joules. Power is work done per unit time, so Power = Work / time = 4900J / 5s = 980 Watts. Therefore, the power rating of the machine is 980 Watts.
What power is used by a machine to perform 800 joules of work in 25 seconds?
Power = energy/time During those 25 seconds, the machine is doing work at the rate of 800/25 = 32 watts. We don't know how much power the machine must consume in order to perform work at that rate, but we know it's more than 32 watts.
A machine exerts a 100 N force to the right over a 5.00 meter length in 4.00 seconds. Calculate the power output of this machine.?
The work done by the machine is calculated as force times distance, which is 100 N * 5.00 m = 500 J. Power is then calculated as work divided by time, yielding 500 J / 4.00 s = 125 watts. Therefore, the power output of the machine is 125 watts.
Which of the following situations involves more power 200 j of work done in 20 s or 50 j of work done in 4 s?
The situation that involves more power is when 200 J of work is done in 20 s. Power is calculated as work done divided by time taken, so for the first situation the power is 10 W (200 J / 20 s) compared to 12.5 W for the second situation (50 J / 4 s).
A machine lifted a 1000 n object 40 meters in 15 seconds how much work is done by the machine?
work = f x s. work = 1000 x 40 40000J
How did a 1980's washing machine work?
please answer.
How much is power output of a machine that lifts a 500kg object through a height of 20m in a time of 60s?
The work done by the machine would be the force required to lift the object multiplied by the distance it was lifted, which is 500kg * 9.81 m/s^2 * 20m = 98100 J. The power output would then be work done divided by time taken, which is 98100 J / 60 s = 1635 W or 1.64 kW.
A 50kg crate is lifted to a height of 10 meters if it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
The work done to lift the crate is equal to the gravitational potential energy gained: Work = force x distance = weight x height. Here, Work = 50kg x 9.8m/s^2 x 10m = 4900 Joules. Power is work done per unit time, so Power = Work / time = 4900J / 5s = 980 Watts. Therefore, the power rating of the machine is 980 Watts.
How has the work of a milkman changed sincethe 1940's?
yh
What power is used by a machine to perform 800 joules of work in 25 seconds?
Power = energy/time During those 25 seconds, the machine is doing work at the rate of 800/25 = 32 watts. We don't know how much power the machine must consume in order to perform work at that rate, but we know it's more than 32 watts.
A machine exerts a 100 N force to the right over a 5.00 meter length in 4.00 seconds. Calculate the power output of this machine.?
The work done by the machine is calculated as force times distance, which is 100 N * 5.00 m = 500 J. Power is then calculated as work divided by time, yielding 500 J / 4.00 s = 125 watts. Therefore, the power output of the machine is 125 watts.
What is technical discription?
technical description is about all functioning of a machine that how it work,s and what it can do
Which of the following situations involves more power 200 j of work done in 20 s or 50 j of work done in 4 s?
The situation that involves more power is when 200 J of work is done in 20 s. Power is calculated as work done divided by time taken, so for the first situation the power is 10 W (200 J / 20 s) compared to 12.5 W for the second situation (50 J / 4 s).
What situations involves more power200 J of work in 20 s or 50 J of work done in 4 s?
Both situations have the same power output of 10 watts, as power is calculated as work done divided by time taken. In this case, 200 J of work done in 20 s and 50 J of work done in 4 s both result in a power output of 10 W.
How much work must be done on a 5kg snowborad to increase its speed from 2ms to 4ms?
The work done on the snowboard can be calculated using the work-energy theorem. The work done is equal to the change in kinetic energy of the snowboard. Given the mass of 5kg and initial and final speeds of 2m/s and 4m/s respectively, the work done on the snowboard is 100 Joules.
What is the power rating of a machine if it took 5 seconds to lift a 50kg crate 10 meters?
The power rating of the machine can be calculated using the formula: Power = Work / Time. The work done is equal to the force applied (50kg * 9.8 m/s^2) multiplied by the distance (10 meters). Therefore, the power rating can be calculated as (50kg * 9.8 m/s^2 * 10m) / 5 seconds.
