Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:
1.195 m
2.8375 s
2.21433 rad/s
Explanation:
d = Distance = 2.39 m
N = Number of cycles = 8
t = Time to complete 8 cycles = 22.7 s
Radius would be equal to the distance divided by 2
The radius is 1.195 m
Time period would be given by
Time period of the motion is 2.8375 s
Angular speed is given by
The angular speed of the motion is 2.21433 rad/s
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:
d = vt = (22 m/s)(12 s) = 264 m
For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²
Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m
Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.
Savanna regions developed during the Triassic period. is true
