Si la velocidad es 3 m/s, y ellos quieren saber la distancia despues 2 segundos, necesita que multiplicar 2 y 3.
La respeusta debiera ser 6m.
Answer:
(3) Both extensional as well as compressional strain is produced
Explanation:
Answer:
Explanation:
We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ
Applying law of conservation of momentum along direction of original motion
m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄
0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ
v cos θ = .8
Applying law of conservation of momentum along direction perpendicular to direction of original motion
1.03 sin 43 x .143 = .132 x v sinθ
v sinθ = .76
squaring and adding
v² = .76 ² + .8²
v = 1.1 m /s
Tan θ = .76 / .8
θ = 44°
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
