When the car starts moving it acquires kinetic energy.
For this problem the formula of kinetic energy is:
![\frac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Where m is the mass of the car = 850kg, and v is the speed of the car.
If we consider insignificant the energy lost by friction with soil and air we can propose the following equation:
Applied energy = Kinetic energy acquired by the car.
![13000\ J = \frac{1}{2}mv^2\\\\13000\ J = \frac{1}{2}(850)v^2\\\\\frac{13000(2)}{850} = v^2\\\\v = \sqrt{\frac{13000(2)}{850}}\\\\v = 5.531 m/s](https://tex.z-dn.net/?f=13000%5C%20J%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5C%5C13000%5C%20J%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28850%29v%5E2%5C%5C%5C%5C%5Cfrac%7B13000%282%29%7D%7B850%7D%20%3D%20v%5E2%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B13000%282%29%7D%7B850%7D%7D%5C%5C%5C%5Cv%20%3D%205.531%20m%2Fs)
The velocity is 5.531 m/s
Hi Pupil Here's Your answer :::
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When an object is moving with uniform circular motion, the centripetal acceleration of the object
![\textbf{is\:Zero}](https://tex.z-dn.net/?f=%5Ctextbf%7Bis%5C%3AZero%7D)
.
Because the body is moving with a uniform velocity and hence, there might not be any acceleration due to it.
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Hope this helps .......
A stationary charge is located between the poles of a horseshoe magnet. The magnetic force exerted by the charge is zero.
<h3>What is charge?</h3>
Charge is the physical property of matter which cause a particle to attract or repel when placed in its field.
A stationary charged particle does not interact with a static magnetic field. A charge placed in a magnetic field experiences a magnetic force. There will be no magnetic force acting on a stationary charge. The charge must be moving in order to have magnetic force on it.
Thus, the magnetic force exerted by the charge is zero.
Learn more about charge.
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The answer is A. wood
because there are three different courts: clay court, grass court, and hard court. wood isn't in there so that would be the answer.
Answer:
which pic...? there is no picture attached to your question