Question

A ship of mass 3x10power7 kg initially at rest is pulled by a force of 5 x 10power4 N through a distance of 3 m.assuming that the resistance due to water is negligible what will be the speed of that ship

Answer 1

F=ma

¶¶ 5*10⁴N=3*10^7kg*acceleration

¶¶ So acceleration of the ship is (5/3)10^-3.

¶¶ Initial velocity of ship is 0

¶¶ Applying 3rd eqn of motion,
¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦V²=0+2*3*(5/3)10^-3¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦

¶¶¶¶¶¶¶¶ And velocity=(1/10)m/s¶¶¶¶¶¶¶¶¶¶¶

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Answer 2

$\huge{\bf{\underline{\underline{\purple{Answer :}}}}}$

Given,

• Mass = $\bold{ 3 \ x \ 10^{7} kg}$
• Initial Velocity - $\bold{0}$
• Force - $\bold{ 5 \ x \ 10^{4} N}$
• Distance [s] - $\bold{ 3 \ m }$
• Final Velocity - ?

We Know this equation : F = ma

From this equation we find Acceleration :

∴ $\implies \bold{\huge{a = \bold{ \frac{F}{m}}}}$

$\implies \bold{\frac{5 \ x \ 10^{4} N }{3 \ x \ 10^{7} Kg }}$

We know this equation also so put the value of Acceleration in this equation :

$\implies \bold{\boxed{ v^{2} = u^{2} \ + \ 2as}}$

$\implies \bold{ v^{2} = 0 + 2 \ x \ \frac{5 \ x \ 10^{4} }{3 \ x \ 10^{7} } \ x \ 3}$

$\implies \bold{ v^{2} = \frac{1}{100}}$

$\implies$ v = 0.1 m\s ................... Answer