The magnitude of the force associated with the gravitational field is constant and has a value f. A particle is launched from po
1 answer:
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Answer:
a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N
Explanation:
The impulse is equal to the variation in the amount of movement.
I =∫ F dt = Δp
I = m - m v₀
Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero
² = V₀² - 2g y
² = 0 - 2 9.8 30.0
= √588
= 24.25 m/s
a) We calculate the impulse
I = 94 24.25 - 0
I = 2279.5 N s
b) Let's join the other expression of the impulse to calculate the average force
I = F t
F = I / t
F = 2279.5 / 6 10⁻³
F = 3.80 10⁵ N
just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n
c) I = m - m v₀
I = 94 8 - 94 (-24.25)
I = 3125.5 N s
d) F = I / t
F = 3125.5 / 6 10⁻³
F = 5.21 10⁵ N
Answer:
Miller Indices are [2, 4, 3]
Solution:
As per the question:
Lattice Constant, C =
Intercepts along the three axes:
Now,
Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.
Now, for the Miller Indices along the three axes:
a =
b =
c =
To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:
a' =
b' =
c' =