0
What else can I help you with?
How do you find the increase in kinetic energy if the percent increase in momentum is given?
Momentum = (mass) x (velocity), which is directly proportional to both mass and velocity.Since mass is constant, any change in momentum is the result of a change in velocity only.If the percent increase 'P' in momentum is given, velocity must have increased to (1 + 0.01P) of its original value.====================Kinetic energy = 1/2 (mass) x (velocity)2, which is directly proportional to mass and to the square of velocity.Since mass is constant, any change in kinetic energy is the result of a change in velocity only.If the velocity changes from its original value by a factor of (1 + 0.01P), the KE changes by a factor of (1 + 0.01P)2.The new KE is (1 + 0.01P)2 or [ 1 + 0.02P + 0.0001P2 ] times its original value.
If percent error in measurement of mass is 3 percent and percent error in measurement of velocity is 4 percent then what will the percentage error in measurement of kinetic energy?
Kinetic Energy = 1/2 (mass) (velocity)2Measurement of mass is in error by 3%.Measurement of velocity is in error by 4%.If both are low, then KE is measured as(True KE) x (.97) x (.96)2 = 0.894 TKE = 10.6% low.If both are high, then KE is measured as(True KE) x (1.03) x (1.04)2 = 1.114 TKE = 11.4% high.If one is high and the other low, then the net error is in between these limits.
What would be the new velocity of a body to make its kinetic energy double?
Momentum = mass * velocity = force * time Kinetic Energy = 1/2 * mass * velocity squared Since the mass of your object is not likely changing, it is the velocity which is changing. Therefore, your new velocity is 1.5 times your initial velocity. In the kinetic energy equation, velocity is squared. 1.5 squared = 2.25, and since nothing else is changing, your kinetic energy is now 2.25 times initial. This is an increase of 125%.
What is the effiency of a motor that converts100j of electrical energy into 70j of useful kinetic energy?
The efficiency of the motor can be calculated as the ratio of useful output energy to input energy. In this case, the efficiency would be 70J (useful kinetic energy) divided by 100J (electrical energy) multiplied by 100% to get a percentage. So, the efficiency of the motor is 70%.
How much energy in joules does a regenerative brake produce when going from 60mph or 25 meter per second to zero?
You can use the formula for kinetic energy (1/2 x mass x velocity squared). However, I don't know how efficient those brakes are; that is, how much percent of the energy typically gets recovered.
If kinetic energy is increased by 60 percent then momentum will be?
Kinetic energy is proportional to the square of the speed; use this fact to calculate the increase in speed (60% increase means an increase by a factor of 1.6). Momentum is proportional to the speed.
How do you find the increase in kinetic energy if the percent increase in momentum is given?
Momentum = (mass) x (velocity), which is directly proportional to both mass and velocity.Since mass is constant, any change in momentum is the result of a change in velocity only.If the percent increase 'P' in momentum is given, velocity must have increased to (1 + 0.01P) of its original value.====================Kinetic energy = 1/2 (mass) x (velocity)2, which is directly proportional to mass and to the square of velocity.Since mass is constant, any change in kinetic energy is the result of a change in velocity only.If the velocity changes from its original value by a factor of (1 + 0.01P), the KE changes by a factor of (1 + 0.01P)2.The new KE is (1 + 0.01P)2 or [ 1 + 0.02P + 0.0001P2 ] times its original value.
If percent error in measurement of mass is 3 percent and percent error in measurement of velocity is 4 percent then what will the percentage error in measurement of kinetic energy?
Kinetic Energy = 1/2 (mass) (velocity)2Measurement of mass is in error by 3%.Measurement of velocity is in error by 4%.If both are low, then KE is measured as(True KE) x (.97) x (.96)2 = 0.894 TKE = 10.6% low.If both are high, then KE is measured as(True KE) x (1.03) x (1.04)2 = 1.114 TKE = 11.4% high.If one is high and the other low, then the net error is in between these limits.
How many is 1 percent in meters?
That depends what the measurement is one percent of.
Which property of a measurement is best estimated from the percent error?
Measurement error: obviously!
What s coefficient of curvature in soil mechanics?
Cc = (D30)^2 / (D10 * D60) Where: * D60 is the particle-size diameter for which 60 percent of the sample was finer, and, * D30 is the particle-size diameter for which 30 percent of the sample was finer. * D10 is the particle-size diameter for which 10 percent of the sample was finer.
What is the angle measurement for 17.5 percent?
14
What would be the new velocity of a body to make its kinetic energy double?
Momentum = mass * velocity = force * time Kinetic Energy = 1/2 * mass * velocity squared Since the mass of your object is not likely changing, it is the velocity which is changing. Therefore, your new velocity is 1.5 times your initial velocity. In the kinetic energy equation, velocity is squared. 1.5 squared = 2.25, and since nothing else is changing, your kinetic energy is now 2.25 times initial. This is an increase of 125%.
If KE increases 300 percent then momentum increased by how many percent?
Assuming the mass isn't changing, 100%. Let the initial velocity be v0 and the mass of the object be m. Then KE0= (1/2)mv02 and P=mv. The new kinetic energy is KE1=4*KE0. (A 300% increase means the change is 3 times the original, giving a new value of 3KE0+KE0=4KE0.) So 4*(1/2)mv02=(1/2)mv12. Simplifying by canceling 1/2 and m, we get 4*v02=v12. Taking the square root of each side: 2*v0=v1. So the new velocity is twice the original. Plugging this into the formula for momentum, we see that the new momentum, P1=mv1=m*2*v0. So the momentum is twice as large as it was initially. The change in momentum is P0, so the percent change is P0/P0, or 100%.
Why does a Percent correct on a memory task is an example of a variable with a scale of measurement and is a numerical observation.?
Why does a Percent correct on a memory task is an example of a variable with a scale of measurement and is a numerical observation.?
What should the percent error be in order to accurately assess the level of precision in the measurement?
The percent error should be as close to zero as possible in order to accurately assess the level of precision in the measurement.
Percent error formula?
Divide the calculated or estimated error by the magnitude of the measurement. Take the absolute value of the result, that is, if it is negative, convert to positive. This would make the percent error = | error / measurement |.
