Question

A force of 50N is applied on a box
of mass 75g. What is the
acceleration produced in it?​

Answer 1

F=ma

50=75/1000×a

=50×1000/75=a

=2000/3=a

=666.6=a

Answer 2

Answer:

Answer:✌✌Hello siso ✌✌

Answer:✌✌Hello siso ✌✌Explanation:

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.As the frictional force of 10 N acts in a direction opposite to movement of box,

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.As the frictional force of 10 N acts in a direction opposite to movement of box,[math]F_{friction}=10 N[/math]

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.As the frictional force of 10 N acts in a direction opposite to movement of box,[math]F_{friction}=10 N[/math]Therefore,[math]F_{net}=F_{applied}-F_{friction}=50-10=40 N[/math]

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.As the frictional force of 10 N acts in a direction opposite to movement of box,[math]F_{friction}=10 N[/math]Therefore,[math]F_{net}=F_{applied}-F_{friction}=50-10=40 N[/math]So,

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.As the frictional force of 10 N acts in a direction opposite to movement of box,[math]F_{friction}=10 N[/math]Therefore,[math]F_{net}=F_{applied}-F_{friction}=50-10=40 N[/math]So,[math]a_{net}=\dfrac{F_{net}}{m} =\dfrac{40}{75}\approx 0.53 m/s^2[/math]

Answer:✌✌Hello siso ✌✌Explanation:We know,F[math]_{net}=ma_{net}.[/math]Here,we are applying a force of 50N,but a frictional force of 10N opposes the movement.As the frictional force of 10 N acts in a direction opposite to movement of box,[math]F_{friction}=10 N[/math]Therefore,[math]F_{net}=F_{applied}-F_{friction}=50-10=40 N[/math]So,[math]a_{net}=\dfrac{F_{net}}{m} =\dfrac{40}{75}\approx 0.53 m/s^2[/math]❤❤BTS ARMY FOREVER ❤❤