Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.50 kg mass to the
1 answer:
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Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
Answer:
a) The average velocity is 32 m/s
b) See the attached figure. Slope of the line = 32.
Graphic: a line that passes through the points (0; 50) and (5; 210)
Explanation:
a) The average velocity is calculated as the displacement over time:
v = ΔX / Δt
where
ΔX : displacement ( final position - initial position)
Δt : time (final time - initial time)
Considering the origin of the reference system as the position where the observer is:
ΔX = 210 m - 50 m = 160 m
Δt = 5 s
v = 160 m / 5 s = 32 m/s
b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:
slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.
