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:
1.5 × 10³⁶ light-years
Explanation:
A certain square region in interstellar space has an area of approximately 2.4 × 10⁷² (light-years)². The area of a square can be calculated using the following expression.
A = l²
where,
A is the area of the square
l is the side of the square
l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years
Answer:
2442.5 Nm
Explanation:
Tension, T = 8.57 x 10^2 N
length of rope, l = 8.17 m
y = 0.524 m
h = 2.99 m
According to diagram
Sin θ = (2.99 - 0.524) / 8.17
Sin θ = 0.3018
θ = 17.6°
So, torque about the base of the tree is
Torque = T x Cos θ x 2.99
Torque = 8.57 x 100 x Cos 17.6° x 2.99
Torque = 2442.5 Nm
thus, the torque is 2442.5 Nm.
Answer:
u = 23.68 m/s
Explanation:
given,
mass of the car, m = 1900 Kg
Force exerted on the wall, F = 150,000 N
time of contact, t = 0.3 s
final speed of the car = 0 m/s
initial speed of the car = ?
we know,
Impulse is equal to change in momentum
J = m v- mu
J = 1900 x 0 - 1900 x (-u)
J = 1900 u
impulse is also equal to force into time
J = F x t
equating both equation of impulse
1900 u = F x t
1900 u = 150000 x 0.3
1900 u = 45000
u = 23.68 m/s
Speed of the car before collision is equal to 23.68 m/s
