The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².
<h3>
Tensile stress of the wire</h3>
The tensile stress of the wire is calculated as follows;
σ = F/A
where;
A = πr² = πD²/4
where;
A = π x (0.64 x 10⁻³)²/4
A = 3.22 x 10⁻⁷ m²
σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)
σ = 6.1 x 10⁷ N/m²
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Answer:
a. slope=rise/run
rise=0.02
run=-2
determined using the point (3,0.08) and (1,0.1) on the graph
slope=0.02/-2
= -0.01 or -1/100
b.area= area of trapizoid+ rectangle
((0.07+0.11)÷2)×4+1×0.07
0.36+0.07
=0.43$
c. the area represent the total cost after 5 hours
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Answer:
F = 104.832 N
Explanation:
given,
upward acceleration of the lift = 1.90 m/s²
mass of box containing new computer = 28 kg.
coefficient of friction = 0.32
magnitude of force = ?
box is moving at constant speed hence acceleration will be zero.
Now force acting due to lift moving upward =
F = μ m ( g + a )
F = 0.32 × 28 × ( 9.8 + 1.9 )
F = 104.832 N
hence, the force applied should be equal to 104.832 N
Answer:
Explanation:
Kinetic energy of block will be converted into heat energy by friction .
Heat energy produced = 1/2 m v²
= .5 x 4.8 x 1.2²
= 3.456 J
85% of energy is converted into heat energy , so heat energy produced
= .85 x 3.456 = 2.9376 J .
If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature
Q = msΔt
specific heat of iron s = 462 J / kg C
Putting the values ,
2.9376 = 4.8 x 462 x Δt
Δt = 13.24 x 10⁻⁴ ⁰C.
