Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.
Acceleration = 1.3 m/s²
Velocity: ∫ 1.3 dx = 1.3x + c m/s
Distance: ∫ 1.3x dx = 1.3x²/2 + c m
Distance run: 1.3*3²/2 = 5.85 m
<em>What</em><em> </em><em>bad</em><em> </em><em>thing</em><em> </em><em>happened</em><em>?</em>
Hi sorry she is a little mean sorry
:( :( :( :( :( :( :( :( :( :( :( :( :( :( :( :( :(
The answer is D. <span>Open
</span>
Explanation:
It is given that,
Weight of the person, W = F = 846 N
When the person steps onto a spring scale in the bathroom, the spring compresses by 0.574 cm, x = 0.00574 m
(a) The force acting on the spring is is given by Hooke's law as :
k = 147386.7 N/m
(b) If the spring is compressed by 0.314 cm or 0.00314 m, weight of the person is given by again Hooke's law as :
F = 462.7 N
Hence, this is the required solution.
The minimum spacing the engineer should leave between the sections to eliminate buckling is 0.011 m.
To calculate the minimum space the engineer must leave in other to eliminate bulking, we use the coefficient of linear expansivity of steel.
<h3>What is coefficient of linear expansivity?</h3>
This can be defined as the increase in length, per unit length, per degree rise in temperature.
To calculate the minimum space, we use the fomula below
Formula:
- α = ΔL/L₁(t₂-t₁)........... Equation 1
Make L₂ the subject of the equation
- ΔL = αL₁(t₂-t₁)........... Equation 2
From the question,
Given:
- L₁ = 25 m
- t₁ = 10°C
- t₂ = 50°C
- α = 11×10⁻⁶ /K ( Cofficient of Linear expansion of steel)
Substitute these values into equation 2
- ΔL = [11×10⁻⁶×25×(50-10)]
- ΔL = 0.011 m
Hence, the minimum spacing the engineer should leave between the sections to eliminate buckling is 0.011 m.
Learn more about cofficient of linear expansion here: brainly.com/question/24215446
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