Answer:
Explanation:
F = ma and
We have F, we have m, but in order to solve for v, we need a.
30.0 = 3.00a so
a = 10.0 m/s/s. Plug that in for a in the second equation and solve for v:
so
v = 10.0(3.00) so
v = 30.0 m/s
Answer:
A bicycle on the top of the hill has the highest potential energy, and when the bike goes down, it transfers to kinetic because it is moving
Explanation:
yeah
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]
Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa