<span>This problem asks that we compute the magnitude
of the maximum stress that exists at the tip of an internal crack
</span> sigmam=2*(sigma0)(a/(pt))^(1/2)
sigmam= 2*(180)*(((3.5×10^-2)/(2))/(4.5×10-4))^(1/2)=2245 MPA (325609.7 PSI)
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
Answer:
Explanation:
A car travels 6.0 km to the north and then 8.0 km to the east. The intensity of the position vector, in relation to the starting point is: a) 14 km b) 2.0 km c) 12 km d) 10 km e) 8.0 km
Check attachment for diagram
The intensity of the position vector is equal to the displacement,
So, to calculate the displacement, we need to find the length of the straight line from starting point to end point.
So, applying Pythagorean theorem
c² = a² + b²
R² = 6² + 36²
R² = 36 + 64
R² = 100
R = √100
R = 10 km.
Verifique el adjunto para ver el diagrama
La intensidad del vector de posición es igual al desplazamiento,
Entonces, para calcular el desplazamiento, necesitamos encontrar la longitud de la línea recta desde el punto inicial hasta el punto final.
Entonces, aplicando el teorema de Pitágoras
c² = a² + b²
R² = 6² + 36²
R² = 36 + 64
R² = 100
R = √100
R = 10 km.
Im not exactly sure but I think the answer is techtonic plates collide
Answer:
0.031 m
Explanation:
Density of copper = ρ = 8960 kg/m³
r = Radius = 0.262 m
m = Mass of plate = 61.5 kg
v = Volume of plate = Volume of cylinder = πr²h

So, thickness of plate is 0.031 m