Answer:
128.21 m
Explanation:
The following data were obtained from the question:
Initial temperature (θ₁) = 4 °C
Final temperature (θ₂) = 43 °C
Change in length (ΔL) = 8.5 cm
Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)
Original length (L₁) =.?
The original length can be obtained as follow:
α = ΔL / L₁(θ₂ – θ₁)
17×10¯⁶ = 8.5 / L₁(43 – 4)
17×10¯⁶ = 8.5 / L₁(39)
17×10¯⁶ = 8.5 / 39L₁
Cross multiply
17×10¯⁶ × 39L₁ = 8.5
6.63×10¯⁴ L₁ = 8.5
Divide both side by 6.63×10¯⁴
L₁ = 8.5 / 6.63×10¯⁴
L₁ = 12820.51 cm
Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
12820.51 cm = 12820.51 cm × 1 m / 100 cm
12820.51 cm = 128.21 m
Thus, the original length of the wire is 128.21 m
Answer:
Explanation:
Assuming school is at the end of the 20 mile route, then
20 mi / 35 mi/hr = 0.57142...hr
which is about 34 minutes 17 seconds
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
Answer:
Put water at room temperature into a vacuum chamber and begin removing the air. Eventually, the boiling temperature will fall below the water temperature and boiling will begin without heating. Or if you want to be easy but messy, add dry ice to a bowl of water and watch how the water starts to boil.
