Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).
Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.
Mathematically it is given by the relation:
Impulse = Force 
Time
According to the numerical values given in the question, I = 202 Ns and T = 0.244 s
So, Force F = 
= 
= 827.86 N
Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).
Answer:
t = 22.2 s
Explanation:
angular distance covered in the 36.0 s is
θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians
720/2 = 360 radians
α = Δω/t = (30 - 10)/36 = 5/9 rad/s²
θ = ω₀t + ½αt²
360 = 10.0t + ½(5/9)t²
0 = (5/18)t² + 10.0t - 360
0 = t² + 36t - 1296
t = (-36 ±√(36² - 4(1)(-1296))) / 2
t = (-36 ±√(6480)) / 2
t = -18 ±√1620
we ignore the negative time result as it occurs before we care.
t = -18 + √1620 = 22.249223... s
Answer:
The input force that you use on an inclined plane is the force with which you push or pull an object. The output force is the force that you would need to lift the object without the inclined plane. This force is equal to the weight of the object.
Explanation:
Answer:
the height reached is = 0.458 [m]
Explanation:
We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:
![Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3AEk%3D%20kinetic%20energy%20%5BJ%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Cv%20%3D%20velocity%20of%20the%20ball%20%5Bm%2Fs%5D)
Replacing the values on the equation we have:
![Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%2A%282%29%2A%283%5E%7B2%7D%20%29%5C%5C%20Ek%3D9%5BJ%5D%5C%5C)
This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.
![Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\](https://tex.z-dn.net/?f=Ek%3DEp%5C%5Cwhere%5C%5CEp%3Dpotential%20energy%20%5BJ%5D%5C%5CEp%3Dm%2Ag%2Ah%5C%5Cwhere%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%3Dheight%20reached%20%5Bm%5D%5C%5C)
Now we have:
![h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]](https://tex.z-dn.net/?f=h%3D%5Cfrac%7BEp%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B9%7D%7B2%2A9.81%7D%20%5C%5C%5C%5Ch%3D0.45%20%5Bm%5D)
In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]
