Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.
<h3>Kinetic energy</h3>
Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.
Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.
The kinetic energy is represented by the following expression:
Ec= ½ mv²
Where:
- Ec is the kinetic energy, which is measured in Joules (J).
- m is the mass measured in kilograms (kg).
- v is the speed measured in meters over seconds (m/s).
<h3>Kinetic energy of a bullet</h3>
In this case, you know:
Replacing in the definition of kinetic energy:
Ec= ½ ×0.500 kg× (25 m/s)²
Solving:
<u><em>Ec= 156.25 J</em></u>
Finally, the bullet has a kinetic energy of 156.25 J.
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Answer:
Ff = 839.05 N
Explanation:
We can use the equation:
Ff = μ*N
where <em>N</em> can be obtained as follows:
∑ Fc = m*ac ⇒ N - F = m*ac = m*ω²*R ⇒ N = F + m*ω²*R
then if
F = 32 N
m = 133 Kg
R = 0.635 m
ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s
we get
N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N
Finally
Ff = μ*N = 0.10*(8390.53 N) = 839.05 N
Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
C Occupational Safety and Health Administration
