If the unit of force and length each increased by 4 times , then the unit of energy is increased by
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Answer: 16 times.
Dimensionally, Force is represented as [MLT−2][MLT−2] and Length as [L][L].
⇒⇒ Energy =Fd=Fd is represented as [MLT−2]×[L]=[ML2T−2][MLT−2]×[L]=[ML2T−2].
Therefore, if Force and length are quadrupled in value, Energy =4[MLT−2]×4[L]=16[ML2T−2]=4[MLT−2]×4[L]=16[ML2T−2].
⇒⇒ the unit of energy increased 16 times.
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Answer: 16 times.
Dimensionally, Force is represented as [MLT−2][MLT−2] and Length as [L][L].
⇒⇒ Energy =Fd=Fd is represented as [MLT−2]×[L]=[ML2T−2][MLT−2]×[L]=[ML2T−2].
Therefore, if Force and length are quadrupled in value, Energy =4[MLT−2]×4[L]=16[ML2T−2]=4[MLT−2]×4[L]=16[ML2T−2].
⇒⇒ the unit of energy increased 16 times.
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Answered by
1
Answer : 16 times
Dimensionally, Force is represented as [MLT−2][MLT−2] and Length as [L][L].
⇒⇒ Energy =Fd=Fd is represented as [MLT−2]×[L]=[ML2T−2][MLT−2]×[L]=[ML2T−2].
Therefore, if Force and length are quadrupled in value, Energy =4[MLT−2]×4[L]=16[ML2T−2]=4[MLT−2]×4[L]=16[ML2T−2].
⇒⇒ the unit of energy increased 16 times.
OR
Let original energy be =E1=f.s
Then, new energy=E2=4f.4s=16f.s=16E1
Hence, energy is increased by 16 times.
OR
f.s=4×4= 16
Dimensionally, Force is represented as [MLT−2][MLT−2] and Length as [L][L].
⇒⇒ Energy =Fd=Fd is represented as [MLT−2]×[L]=[ML2T−2][MLT−2]×[L]=[ML2T−2].
Therefore, if Force and length are quadrupled in value, Energy =4[MLT−2]×4[L]=16[ML2T−2]=4[MLT−2]×4[L]=16[ML2T−2].
⇒⇒ the unit of energy increased 16 times.
OR
Let original energy be =E1=f.s
Then, new energy=E2=4f.4s=16f.s=16E1
Hence, energy is increased by 16 times.
OR
f.s=4×4= 16
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