\Delta L= \alpha L_0 (T_f-T_i)
= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.
Run inside if you are outdoors
.
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
Answer: <span>D. A bimetallic strip bends so that the steel is on the outside curve
</span>
When something has an increased temperature, its volume will expand. Then, if the temperature drops, its volume should be smaller. From there option A and B are out since the liquid in thermometer is expand or move up.
When you put two kinds of different metal with a different coefficient of thermal expansion, the outer curve metal will be the one with lesser coefficient when temperature drop. Since the question about drop in temperature then the metal should be bend
Brass will expand 1.5 times more than the steel so the outer curve would be the steel.
Answer:
Average speed = 1.2 m/s
Average velocity = 0.4 m/s
Explanation:
Average speed = total distance/total time
Average speed = (40 + 20)/(40 + 10)
Average speed = 60/50
Average speed = 1.2 m/s
Average velocity = displacement/time
Now, she ran 40 m south and ran 20 m back north which is in the direction of where she began the journey.
Thus;
Displacement = 40 - 20 = 20 m
Average velocity = 20/50 = 0.4 m/s
