2017年10月11日 · The minimum work required to push a 950-kg car up a 9.0° incline for 710 meters is approximately 1.03 million joules. This is calculated using the gravitational force …

What is the minimum work needed to push a 950-kg car 710 m up along a 9.0 ^\circ 9.0∘ incline? Ignore friction. In order to watch this solution you need to have a subscription. This is Giancoli …

Here, F is the force required to bring the car into motion, m is the mass of car (m = 950 k g), v is the final velocity of the car (v = 90 k m / h), u is the initial velocity of the car (u = 0 as the car …

We are aware that it takes energy to get an object, like a car or the package in Figure 2, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This …

Minimum work, in this case, is the product of the weight of the car, W = 950 kg * 9.81 m'/s^2, by the height 710 m minimum work = 950*9.81*710 = 6616845 J (joules). ANSWER It DOES …

2018年10月31日 · The average force required to stop a 950 kg car traveling at 95 km/h in 8 seconds is approximately -3135 N. This negative value indicates that the force is acting in the …

What is the minimum work needed to push a 950-kg car 710 m up along a 9.0 degree incline? Ignore friction. GIVENS (lowest number of digits: 2, checked International system of units) …

We're going to calculate the force needed to bring a 950 kilogram car to a stop over a distance of 120 meters, assuming an initial velocity of 90 kilometers per hour.

2023年11月4日 · Consider all the forces acting on the car in the horizontal and vertical directions. The minimum work is done on the car when it is moving with steady velocity. Given data: The …

Next, we can use the formula for gravitational potential energy to find the minimum work needed: potential energy = mass x gravity x height potential energy = 950 kg x 9.8 m/s^2 x 103.9 m …