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Question: Convert 1 MW power on a new system having basic units of mass, length and time as 10 kg, 1 dm and 1 ...

Convert 1 MW power on a new system having basic units of mass, length and time as 10 kg, 1 dm and 1 min respectively.

Explanation

Solution

The above problem is based on the conversion of the units. The units used for the measurement of physical quantities may be in MKS system, CGS system or FPS system. The watt is the unit of measurement of the power. It is the same as the work done per second. The work done is the same as the displacement of the object due to application of the force.

Complete step by step answer:
The power of the system is 1 MW, mass for the conversion is 10 kg, displacement of the object is 1 dm and time is 1 min.

1 MW can be expressed as in the basic units :
1  MW=1×106  J/s......(1)1\;{\text{MW}} = 1 \times {10^6}\;{\text{J}}/{\text{s}}......\left( 1 \right)
1 Joule can be expressed as:
1  J=kg⋅m2/s21\;{\text{J}} = {\text{kg}} \cdot {{\text{m}}^2}/{{\text{s}}^2}
1 second can be expressed in minutes as:
1  min⁡=60  s1\;\min = 60\;{\text{s}}
1  s=160  min⁡1\;{\text{s}} = \dfrac{1}{{60}}\;\min
1 m can be expressed as:
1  dm=10  m1\;{\text{dm}} = 10\;{\text{m}}
1  m=110  dm1\;{\text{m}} = \dfrac{1}{{10}}\;{\text{dm}}

Substitute all the basic conversions in the expression (1).
1  MW=1×106  kg⋅m2/s2s1\;{\text{MW}} = \dfrac{{1 \times {{10}^6}\;{\text{kg}} \cdot {{\text{m}}^2}/{{\text{s}}^2}}}{{\text{s}}}
⇒1  MW=1×106  kg⋅m2s3\Rightarrow 1\;{\text{MW}} = \dfrac{{1 \times {{10}^6}\;{\text{kg}} \cdot {{\text{m}}^2}}}{{{{\text{s}}^3}}}
⇒1  MW=1×106  kg⋅(110  dm)2(160  min⁡)3\Rightarrow 1\;{\text{MW}} = \dfrac{{1 \times {{10}^6}\;{\text{kg}} \cdot {{\left( {\dfrac{1}{{10}}\;{\text{dm}}} \right)}^2}}}{{{{\left( {\dfrac{1}{{60}}\;\min } \right)}^3}}}
⇒1  MW=216×1011  kg⋅dm2/min⁡3\Rightarrow 1\;{\text{MW}} = 216 \times {10^{11}}\;{\text{kg}} \cdot {\text{d}}{{\text{m}}^2}/{\min ^3}

Therefore, the 1 MW power on a new system having basic units of mass, length and time is 216×1011  kgdm2/min⁡3216 \times {10^{11}}\;{\text{kg}}{\text{d}}{{\text{m}}^2}/{\min ^3}.

Note: Use the conversion very carefully as the variation in the power of any unit can cause the accuracy in the conversion. The above conversion can also be done by using the dimensional formula. The dimensional formula describes the physical quantities used in the formula.