Answer:
a car
A sled sliding across snow or ice.
a ball down a hill
mercury
Explanation:
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:
ω = 3.1 rad/s
θ = 36° from vertical
Explanation:
I will ASSUME that the bob and string is acting as a pendulum.
Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.
At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.
F = mg + mω²R
F/m = g + ω²R
F/m - g = ω²R
ω = √((F/m - g)/R)
ω = √((3/0.220 - 9.8)/0.40)
ω = 3.09691...
ω = 3.1 rad/s
Potential energy will convert to kinetic energy
mgh = ½mv²
h = v²/2g
R - Rcosθ = v²/2g
R(1 - cosθ) = v²/2g
1 - cosθ = v²/2gR
cosθ = 1 - v²/2gR
cosθ = 1 - (Rω)²/2gR
cosθ = 1 - Rω²/2g
cosθ = 1 - 0.40(3.1²)/(2(9.8))
cosθ = 0.804267
θ = 36.46045...
θ = 36°
I think it should be a cold
