answersLogoWhite

0

The work done is equal to the change in kinetic energy, which can be calculated using the formula: W = ΔKE = 1/2 m (v_f^2 - v_i^2). Plugging in the values, the work done to increase the speed of the scooter from 10 m/s to 20 m/s is 6000 J.

User Avatar

AnswerBot

1y ago

What else can I help you with?

Continue Learning about Physics

How much work must be done on a kg sled to increase its speed from 10ms to 15 ms?

To calculate the work done on the sled to increase its speed, you need to know the initial and final kinetic energy. The work done is equal to the change in kinetic energy, which is given by the formula: Work = (1/2) * m * (vf^2 - vi^2) Substitute the mass of the sled, initial speed, and final speed to find the work done.


How much work is done to increase a 20 kg scooter speed from Ms to 20 ms?

The work done to increase the speed of the scooter can be calculated as the change in kinetic energy. The initial kinetic energy is given by (1/2)mv^2. The final kinetic energy is (1/2)m(20)^2. The work done is the difference between the final and initial kinetic energies.


How much work must be done on a 10-kg snowboard to increase its speed from 2ms to 4 ms 100 J.?

The work done on the snowboard to increase its speed is 100 Joules. This work-energy theorem is based on the change in kinetic energy, given by the equation: work = change in kinetic energy. In this case, the snowboard has a mass of 10 kg and the speed increases from 2 m/s to 4 m/s.


How much work must be done on a 1000-kg car to increase its speed from 1 ms to 2 ms?

The work done to increase the speed of an object is equal to the change in its kinetic energy. The change in kinetic energy can be calculated using the formula ΔKE = 0.5 * m * (vf^2 - vi^2), where m is the mass of the object, vf is the final velocity, and vi is the initial velocity. Substituting the given values, the work done on a 1000-kg car to increase its speed from 1 m/s to 2 m/s would be ΔKE = 0.5 * 1000 * (2^2 - 1^2) = 1000 J.


How much work must be done on a kg shopping cart to increase its speed from 1 ms to 4 ms?

Probably less then solving this question or since F=ma 72N. The N stands for Nitwick.

Related Questions

How much work is done to increase a 20 kg scooters speed from 10 ms to 20 ms?

3000j


How much work must be done on a kg sled to increase its speed from 10ms to 15 ms?

To calculate the work done on the sled to increase its speed, you need to know the initial and final kinetic energy. The work done is equal to the change in kinetic energy, which is given by the formula: Work = (1/2) * m * (vf^2 - vi^2) Substitute the mass of the sled, initial speed, and final speed to find the work done.


Are slam scooters the same as pro stunt scooters?

I have owned both of these scooters and the pro scooters are much better, That's why they cost more than slam scooters.


What is the speed restriction on 50cc scooters?

50cc scooters are restricted to around 30mph here in the UK. They can be de-reg'd up by a few mph but you'll notgenerally get much more out of them. Weight and stuff are also factors.


How much work must be done on a 5-kg snowboard to increase its speed from 2 m s to 4 m s?

30 J


How much work must be done on a 5-kg sled to increase its speed from 10 m/s to 15 m/s?

312.5 J


How much work must be done on a 10-kg bicycle to increase its speed from 5 m s to 10 m s?

375 Js (((((((((((: this is the right answer


How much work is done to increase a 20-kg scooter's speed from 10 m/s to 20 m/s?

3000 J *Shelby Sarah*


How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?

750 j


How much work must be done on a 10kg bicycle to increase its speed from 2 M's to 3 M's?

Work done = increase in kinetic energy ie 1/2 * 10 * (3+2)(3-2) [recall a2 - b2 = (a+b)(a-b)] Hence work done = 25 joule.


How much work is done to increase a 20 kg scooter speed from Ms to 20 ms?

The work done to increase the speed of the scooter can be calculated as the change in kinetic energy. The initial kinetic energy is given by (1/2)mv^2. The final kinetic energy is (1/2)m(20)^2. The work done is the difference between the final and initial kinetic energies.


How much do scooters cost?

$70