Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
<u>Distance stretched ( Δl ) = 4 * 10^-3 m </u> ( right value )
<u>average bond length ( between atoms ) = 2.3 * 10^-10 m </u>( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
The final velocity of the object is 16m\s.
Hope this helps! :)
Answer:
please find your answer in the attached picture, along with explanation.
Answer:
The force required to hold the contraction in place is 665.91 N ↑
Explanation:
Given;
specific gravity of oil, γ = 0.75
Volumetric flow rate, V 3.2 Ft³/s = 0.0906 m³/s
where;
is the density of oil
is the density of water = 1000 kg/m³
∴density of oil () = γ × density of water()
= 0.75 × 1000 kg/m³
= 750kg/m³
Buoyant Force = ρVg
= 750 × 0.0906 × 9.8
= 665.91 N ↑
This force acts upward or opposite gravitational force.
Therefore, the force required to hold the contraction in place is 665.91 N ↑
