Question

Question:( Hindi A body of weight 200 N is placed on a rough horizontal plane. Angle of friction is 20°. Magnitud of the force, which can move the body, while acting in the horizontal direction is​

Answer 1

Explanation:

1)What is the remainder , when X²⁵ - 25 is divided by X + 1?

join

here

hfh-bpju-vjq

Answer 2

Answer:

According to question,

As shown in below figure

So,The force on the block long the horizontal is

F

x

=100cos30−f⋯(i)

f=μ×N⋯(ii)

and also,

200=N+100sin30

N=150N

Now,

substituting the above value in (ii)

f=150×μ

substituting the value of

′

f

′

in (i)

F

x

=100cos30−150×μ

Hence, the block moves with the constant velocity it means that there is no net force acting on it is horizontal direction,

F

x

=0

100cos30−150×μ=0

86.6025=150×μ

μ=0.57735

Explanation:

Hope this helps you