Solid particles hope this helped!
<span>The answer is, 7.44 kg*m/s</span>
Average speed = distance travelled / time used
When someone is holding something that has been struck or splashed by lightning, contact damage occurs.
We need additional information concerning lightning and injuries in order to identify the solution.
<h3>What types of injuries are brought on by lightning?</h3>
- Lightning is the name for a natural electrical discharge that occurs quickly and with a dazzling flash.
- It has a tremendous amount of energy.
- Lightning-related injuries can be divided into three categories: direct strikes, side splashes, and contact injuries.
- When someone is struck by lightning directly, they can get direct injury.
- When a current splashes from a neighboring object, it is called a side splash.
- When someone touches a lightning-hit object, contact harm results.
In light of this, we can say that contact injuries happen when a person is holding an object that has been struck by lightning or splashed by it.
Learn more about the lightning and harm here:
brainly.com/question/28055828
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Answer:
The high of the ramp is 2.81[m]
Explanation:
This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.
If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.
We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.
![E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%5C%5Cwhere%3A%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5Cv%20%3D%202.8%5Bm%2Fs%5D%5C%5Cm%3D%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2Av%5E%7B2%7D%20%7D%20%5C%5Cm%3D%5Cfrac%7B3.8%7D%7B0.5%2A2.8%5E%7B2%7D%20%7D%20%5C%5Cm%3D0.969%5Bkg%5D)
Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.
![E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%2BW_%7Bf%7D%3DE_%7Bk%7D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5CW_%7Bf%7D%3D23%5BJ%5D%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5C)
And therefore
![m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%20%2B%20W_%7Bf%7D%3D3.8%5C%5C%200.969%2A9.81%2Ah%20-%2023%3D%203.8%5C%5Ch%20%3D%20%5Cfrac%7B23%2B3.8%7D%7B0.969%2A9.81%7D%5C%5C%20h%20%3D%202.81%5Bm%5D)
