Answer:
1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]
Explanation:
1. We know that the density is defined by the following expression.
![Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5Cwhere%3A%5C%5Cmass%3D90%5Bkg%5D%5C%5Cvolume%3D0.075%5Bm%5E%7B3%7D%20%5D%5C%5Cdensity%3D%5Cfrac%7B90%7D%7B0.075%7D%20%5C%5Cdensity%3D1200%5B%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%20%7D%5D)
2. First we need to convert the units to meters.
wide = 35[cm] = 35/100 = 0.35[m]
long = 11 [dm] = 11 decimeters = 11/10 = 1.1[m]
Thick = 15[mm] = 15/1000 = 0.015[m]
Now we can find the density using the expression for the density.
![density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]](https://tex.z-dn.net/?f=density%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5Cwhere%3A%5C%5Cvolume%20%3D%20wide%2Along%2Athick%5C%5Cvolume%3D0.35%2A1.1%2A0.015%20%3D%200.005775%5Bm%5E3%5D%5C%5C%5C%5Cmass%3D%20density%2Avolume%20%3D%202700%2A0.005775%20%3D%2015.59%5Bkg%5D)
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer:
Potential energy
Explanation:
Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.