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Work is a scalar quantity, not dependent upon direction. Use simple addition (not a vector sum) to calculate the total work done on the cart: 300. J + 400. J = 700. J.
You can't calculate the force from the rate at which work is done.10 joules in 20 seconds could be:-- 1 newton pushing along at 50 centimeters per second, or-- 0.1 newton pushing 5 meters per second, oran infinite number of other possibilities.From the data given in the question, all we know is that the 'power' is 1/2 watt.
How do you calculate the mass of an object that has a potential energy of 180 joules and rest at the top of a hill 15 meters from the ground?
The 400g is a variable that is meant to fool you and doesn't doesn't apply to the formula. We use the W=Fs formula, Where W= Work (joules), F=Force (Newtons) and s=Displacement (meters). W=? F=2.4N s=0.75 m W=Fs W=2.4x0.75 W=1.8 J
1.03 x 10^6 J
700 joules
Well, the first step is to learn joules and you'll be set!
Work is a scalar quantity, not dependent upon direction. Use simple addition (not a vector sum) to calculate the total work done on the cart: 300. J + 400. J = 700. J.
You can't calculate the force from the rate at which work is done.10 joules in 20 seconds could be:-- 1 newton pushing along at 50 centimeters per second, or-- 0.1 newton pushing 5 meters per second, oran infinite number of other possibilities.From the data given in the question, all we know is that the 'power' is 1/2 watt.
How do you calculate the mass of an object that has a potential energy of 180 joules and rest at the top of a hill 15 meters from the ground?
There are 4.184 joules in a calorie, therefore to burn off 464.. 464/4.184 = 110.899 calories.
The energy of this photon is 3,7351.10e-19 joules.
Watts is joules per second, so 50 joules in 5 seconds is 10 watts.
The 400g is a variable that is meant to fool you and doesn't doesn't apply to the formula. We use the W=Fs formula, Where W= Work (joules), F=Force (Newtons) and s=Displacement (meters). W=? F=2.4N s=0.75 m W=Fs W=2.4x0.75 W=1.8 J
You can divide any unit of energy by any unit of time; but the standard SI unit is Joules / second, and has the special name "Watt".
Work done (joules) and time taken (seconds) is the information needed to calculate power in watts (joules/second).
1 Watt = 1 Joule/second 1 kilowatt = 1000 Watts = 1000 Joules/second 1 hour = 60 minutes = 3600 seconds 1 kWh = 1000 (Joules/second) * 3600 (seconds) = 3,600,000 Joules = 3.6 MJoules