a. Define each of the following terms used with simple machines:
i. Pivot ii.Load iii. Efficiency.
b. A truck of mass 1.2 × \(10^3\) kg is pulled from rest by a constant horizontal force of 25.2N on a leveled road. If the maximum speed attainable in the process is 60 km/h.
Calculate the: i. work done by the force; ii. distance traveled by the truck in reaching the maximum speed.
c. State two differences between absolute zero temperature and ice point.
d. An uncalibrated liquid-in-glass thermometer was used in determining a Celsius temperature. The readings are tabulated below
Temperature/°C | -6 | 0 | 100 |
Length of column/ cm | L | 2.0 | 15.0 |
a
i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates.
ii. A load is the object or weight that is being moved, lifted, or supported by the machine.
iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss
b.
i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s
Workdone = ΔK.E = \(K.E_2 - K. E_1\)
Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\)
Workdone = \(\frac{1}{2}m(v^2 - u^2)\)
Workdonk = \(\frac{1}{2}\times 1200 \times 16.67^2\) ( since u = 0)
Workdone = 600 x 277.89 = 166733.34J
Therefore, Workdone = \(1.67 \times 10^5\)J
ii. F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\)
recall, \(v^2 = u^2 +2aS\)
S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\)
S = 6616.4m = 6.62km.
c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium.
- Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15°C while ice point is typically defined as 0°C at sea level
d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\)
\(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\)
6 x 13 = 100( 2 - L )
78 = 200 - 100L
100L = 200 - 78
100L = 122
L = \(\frac{122}{100}\) = 1.22cm
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