2NaOH+2B+2H2O gives 2NaBO2+3H2 chemical name
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Position of the particle x=2t2−t3. Velocity v =dtdx=4t−3t2. Acceleration a=dtdv=4−6t. So, a( at t=2 s) =4−6×2=−8 m/s2. Since, acceleration is depending on the ...
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TEJAS YE MERE STANDARD KE QUESTION NAHI H
M 8 STANDARD ME HU
so I didn't help you sorry e and you will find this answer on tutorix app clearly
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