Class 9 Exam > Class 9 Questions > A pump lifts a pump lifts 420kg of water ever...
Start Learning for Free
A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump?
Most Upvoted Answer
A pump lifts a pump lifts 420kg of water every minute through a vertic...
A) Potential energy = mgh
= 420×10×30
= 126000 J
b) Power= work done/time
=126000/60
= 2100 W...
= 420×10×30
= 126000 J
b) Power= work done/time
=126000/60
= 2100 W...
Community Answer
A pump lifts a pump lifts 420kg of water every minute through a vertic...
Potential Energy Calculation:
The potential energy of an object is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass of the water being lifted is 420 kg, the acceleration due to gravity is 10 N/kg, and the height is 30 m. Plugging these values into the equation, we can calculate the potential energy:
PE = (420 kg) * (10 N/kg) * (30 m)
PE = 126,000 N*m
Therefore, the potential energy of the water being lifted every minute is 126,000 N*m.
Useful Power Output Calculation:
Power is defined as the rate at which work is done or energy is transferred. In this case, the pump is lifting the water through a height of 30 m every minute. The work done by the pump is equal to the potential energy gained by the water. The formula for power is P = W/t, where P is power, W is work, and t is time. Rearranging the formula, we have W = P * t.
To calculate the useful power output of the pump, we need to determine the work done by the pump in one minute. The work done is equal to the potential energy gained by the water, which we calculated to be 126,000 N*m. The time is given as one minute, which is equal to 60 seconds.
Using the formula, we can calculate the power:
P = (126,000 N*m) / (60 s)
P = 2,100 N*m/s
Therefore, the useful power output of the pump is 2,100 N*m/s.
Explanation:
The potential energy of the water being lifted is calculated by multiplying the mass, acceleration due to gravity, and height. This calculation gives us the amount of potential energy gained by the water every minute.
The useful power output of the pump is calculated by dividing the work done (which is equal to the potential energy gained by the water) by the time in seconds. This calculation gives us the rate at which the pump is transferring energy or doing work.
In this case, the pump is lifting 420 kg of water every minute through a height of 30 m. The potential energy gained by the water is 126,000 N*m, and the useful power output of the pump is 2,100 N*m/s. This means that the pump is transferring energy to the water at a rate of 2,100 N*m/s.
The potential energy of an object is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass of the water being lifted is 420 kg, the acceleration due to gravity is 10 N/kg, and the height is 30 m. Plugging these values into the equation, we can calculate the potential energy:
PE = (420 kg) * (10 N/kg) * (30 m)
PE = 126,000 N*m
Therefore, the potential energy of the water being lifted every minute is 126,000 N*m.
Useful Power Output Calculation:
Power is defined as the rate at which work is done or energy is transferred. In this case, the pump is lifting the water through a height of 30 m every minute. The work done by the pump is equal to the potential energy gained by the water. The formula for power is P = W/t, where P is power, W is work, and t is time. Rearranging the formula, we have W = P * t.
To calculate the useful power output of the pump, we need to determine the work done by the pump in one minute. The work done is equal to the potential energy gained by the water, which we calculated to be 126,000 N*m. The time is given as one minute, which is equal to 60 seconds.
Using the formula, we can calculate the power:
P = (126,000 N*m) / (60 s)
P = 2,100 N*m/s
Therefore, the useful power output of the pump is 2,100 N*m/s.
Explanation:
The potential energy of the water being lifted is calculated by multiplying the mass, acceleration due to gravity, and height. This calculation gives us the amount of potential energy gained by the water every minute.
The useful power output of the pump is calculated by dividing the work done (which is equal to the potential energy gained by the water) by the time in seconds. This calculation gives us the rate at which the pump is transferring energy or doing work.
In this case, the pump is lifting 420 kg of water every minute through a height of 30 m. The potential energy gained by the water is 126,000 N*m, and the useful power output of the pump is 2,100 N*m/s. This means that the pump is transferring energy to the water at a rate of 2,100 N*m/s.
Attention Class 9 Students!
To make sure you are not studying endlessly, EduRev has designed Class 9 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 9.
|
Explore Courses for Class 9 exam
|
|
Similar Class 9 Doubts
Top Courses for Class 9View all
A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump?
Question Description
A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump?.
A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump?.
Solutions for A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? in English & in Hindi are available as part of our courses for Class 9.
Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free.
Here you can find the meaning of A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? defined & explained in the simplest way possible. Besides giving the explanation of
A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump?, a detailed solution for A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? has been provided alongside types of A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? theory, EduRev gives you an
ample number of questions to practice A pump lifts a pump lifts 420kg of water every minute through a vertically height of 30m g=10n/kg a) calculate potential energy every minute b) useful power output of pump? tests, examples and also practice Class 9 tests.
|
|
Explore Courses for Class 9 exam
|
|
Suggested Free Tests
Test: The Fundamental Unit of Life- Case Based Type Questions- 1
Test | 10 questions
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup