final velocity = initial
velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration
distance = (initial velocity x time) +
1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
<span>distance = 0.18 m</span></span>
This is True
Kinetic energy is the energy of motion. The bicyclist is in motion as he pedals up the tall hill. Therefore, the bicyclist contains kinetic energy.
Answer:
second is the SI unit of time
Answer:
The normal stress is 10.7[MPa]
Explanation:
The normal stress can be calculated with the following equation:
![S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]](https://tex.z-dn.net/?f=S_%7Bnorm%7D%20%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5Cwhere%3A%5C%5CF%3D%20force%20%5BNewtons%5D%5C%5CA%3Darea%20%5Bm%5E2%5D%5C%5CS_%7Bnorm%7D%20%3D%20Normal%20stress%20%5B%5Cfrac%7BN%7D%7Bm%5E%7B2%7D%20%7D%5D%20or%20%5BPa%5D)
The area of the rod can be calculated using the equation:
![A=\frac{\pi }{4}*d^{2} \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2} \\A=5.02*10^{-5} [m^{2} ]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2Ad%5E%7B2%7D%20%20%5C%5Cd%3D8%5Bmm%5D%3D0.008%5Bm%5D%5C%5CA%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2A%280.008%29%5E%7B2%7D%20%20%5C%5CA%3D5.02%2A10%5E%7B-5%7D%20%5Bm%5E%7B2%7D%20%5D)
The force is the result of the mass multiplied by the gravity.
![F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]](https://tex.z-dn.net/?f=F%3D55%5Bkg%5D%2A9.81%5Bm%2Fs%5E%7B2%7D%20%5D%20%3D%20539.6%5BN%5D%5C%5C%5C%5CS_%7Bnorm%7D%20%3D%20539.6%2F5.02%2A10%5E%7B-5%7D%20%5C%5CS_%7Bnorm%7D%20%3D%2010.7%2A10%5E%7B6%7D%5BPa%5D%20%3D%2010.7%5BMPa%5D)
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
