Answer:
carbon + oxygen → carbon dioxide
Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.
where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
![V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]](https://tex.z-dn.net/?f=V%20%3D%200.15%2A0.002%2A0.1%5C%5CV%20%3D%200.00003%20%5Bm%5E%7B3%7D%20%5D)
And the mass m = 4 [gramm] = 0.004 [kg]
![Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]](https://tex.z-dn.net/?f=Ro%20%3D%200.004%2F0.00003%5C%5CRo%20%3D%20133%20%5Bkg%2Fm%5E%7B3%7D%5D)
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
