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A jet airplane travelling at the speed of 500 kmh-1 ejects the burnt gases at the speed of 1400 kmh-1 relative to the jet airplane. The speed of burnt gases relative to stationary observer on the earth is(A) 2.8 kmh-1(B) 190 kmh-1(C) 700 kmh-1(D) 900 kmh-1 |
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Answer» Correc Option is (D) 900 \(kmh^{-1}\) \(v_p = -500\) km/h \(V_{gp} = 1400\) km/h The speed of burnt gas with respect to the jet airplane is \(\overrightarrow v_{gp} = \overrightarrow v_g - \overrightarrow v_p\) \(1400 = v_g + 500\) \(v_g = 1400 - 500\) \(v_g = 900\) km/h Correct option is: (D) 900 kmh-1 |
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