Answer:
Shearing strain will be 0.1039 radian
Explanation:
We have given change in length 
Length of the pad L = 1.15 inch
We have to find the shearing strain
Shearing strain is given by
Shearing strain is always in radian so we have to change angle in radian
So 
Answer:
Springs store energy when compressed and release energy when they rebound
Explanation:
Answer:
Ano klassing tanong yn?
Explanation:
Ang taas namn yn? Paki linaw po para matulungan po kita.!!
A and C is the answer to the question. Be 15 years old & get a permit
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
