Home
Class 12
PHYSICS
If a ship of mass 4 xx 10 ^( 7...

If a ship of mass ` 4 xx 10 ^( 7 ) ` kg initially at rest is pulled by a force of ` 5 xx 10 ^ ( 4 ) ` N through a distance of 4 m, then the speed of the ship will be (resistance due to water is negligible )

A

1.5 m/sec

B

60m/sec.

C

0.1m/sec.

D

5m/sec.

Text Solution

Verified by Experts

The correct Answer is:
C

`F=ma rArr a=(F)/(m)=(5x10^(4))/(3x10^(7))=(5)/(3) xx10^(-3) ms^(-2)`
Also `v^(2)-u^(2)=2as`
`rArr v^(2)-0^(2)=2xx(5)/(3)xx10^(-3)xx3=10^(-2) rArr v=0.1 ms^(-1)`
Promotional Banner

Topper's Solved these Questions

  • LAWS OF MOTION

    DISHA PUBLICATION|Exercise Exercise-1: Concept Builder (Topicwise) (Topic 2: Momentum, Law of Conservation of Momentum and Impulse)|15 Videos
  • LAWS OF MOTION

    DISHA PUBLICATION|Exercise Exercise-1: Concept Builder (Topicwise) (Topic 3: Equilibrium of Forces, Motion of Connected Bodies and Pulley )|12 Videos
  • LAWS OF MOTION

    DISHA PUBLICATION|Exercise Exercise-2 : Concept Applicator|30 Videos
  • JEE MAINS- 2019 (HELD ON :9TH APRIL 2019 (SHIFT-I))

    DISHA PUBLICATION|Exercise QUESTIONS|30 Videos
  • MAGNETISM AND MATTER

    DISHA PUBLICATION|Exercise EXERCISE-2:CONCEPT APPLICATOR|31 Videos

Similar Questions

Explore conceptually related problems

A ship of mass 3xx10^(2)kg initially at rest is pulled by a force of 5xx10^(4) N through a distance of 3m. Neglecting frcition, the speed of the ship at this moment is:

A ship of mass 3xx10^7kg initially at rest, is pulled by a force of 5xx10^5N through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship is

. The mass of a ship is 2 xx 10^7 kg. On applying a force of 25 xx 10^5 N, it is displaced through 25 m. The speed of ship is

A constant force acts on a body of mass 5 kg at rest for 10s. If the body moves through a distance of 250 m, what is the magnitude of the force?

A proton of mass 1.6 xx 10^(-27)kg goes round in a circular orbit of radius 0.10 m under a centripetal force of 4 xx 10^(-13) N . then the frequency of revolution of the proton is about

3 xx 10^(5) + 4 xx 10^(3) + 7 xx 10^(2) + 5 is equal to